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\displaystyle x\in\left[\log_2\!\left(\frac{\sqrt5-1}{2}\right),\,1\right].
levou 26 s
Como pensou
Como 2^x+2>2, temos \log_2(2^x+2)>0. Assim, a desigualdade equivale a
\log_2(2^x+2)\ge |2x|
\iff
\begin{cases}
\log_2(2^x+2)\ge 2x,\\
\log_2(2^x+2)\ge -2x.
\end{cases}
Fazendo t=2^x>0 e usando que a função 2^u é crescente, obtemos
\begin{cases}
t+2\ge t^2,\\
t+2\ge t^{-2}.
\end{cases}
A primeira desigualdade fornece
t^2-t-2=(t-2)(t+1)\le0.
Como t>0, resulta t\le2.
Na segunda, multiplicando por t^2>0, temos
t^3+2t^2-1=(t+1)(t^2+t-1)\ge0.
Como t+1>0, devemos ter t^2+t-1\ge0. Suas raízes são \frac{-1\pm\sqrt5}{2}; portanto, para t>0,
t\ge\frac{\sqrt5-1}{2}.
Reunindo as condições e voltando a x,
\frac{\sqrt5-1}{2}\le 2^x\le2
\iff
\log_2\!\left(\frac{\sqrt5-1}{2}\right)\le x\le1.
Resposta final: \displaystyle x\in\left[\log_2\!\left(\frac{\sqrt5-1}{2}\right),\,1\right].
Respondeu · mesma resposta final de outras 8 IAs
\displaystyle x\in\left(\log_2\left(\frac{\sqrt5-1}{2}\right),\,1\right).
levou 23 s
Como pensou
Como 2^x+2>2, temos \log_2(2^x+2)>1>0. Assim, a desigualdade equivale a
\log_2(2^x+2)>2|x|.
Caso x\geq 0: pela monotonicidade do logaritmo,
\log_2(2^x+2)>2x
\iff 2^x+2>2^{2x}.
Fazendo t=2^x\geq1, obtemos
t^2-t-2<0
\iff (t-2)(t+1)<0.
Logo, 1\leq t<2, isto é,
0\leq x<1.
Caso x<0: agora,
\log_2(2^x+2)>-2x
\iff 2^x+2>2^{-2x}.
Com t=2^x\in(0,1), segue que
t+2>\frac1{t^2}
\iff t^3+2t^2-1>0
\iff (t+1)(t^2+t-1)>0.
Como t>0, essa condição equivale a
t>\frac{\sqrt5-1}{2}.
Portanto,
\log_2\left(\frac{\sqrt5-1}{2}\right)<x<0.
Unindo os dois casos, obtemos o intervalo procurado.
Resposta final: \displaystyle x\in\left(\log_2\left(\frac{\sqrt5-1}{2}\right),\,1\right).
Respondeu · mesma resposta final de outras 8 IAs
\displaystyle \log_2\left(\frac{\sqrt5-1}{2}\right)<x<1.
levou 13 s
Como pensou
**Checking notation and solutions**
I see that x=0 is included, which is good. I might need to check the log argument notation, like log_2(2^x+2). Okay, that makes sense. I'll want to develop a candidate solution based on that. It feels clear to me that I'm on the right path, just need to ensure all parts are aligned correctly. Let's keep going!
Como 2^x+2>2, temos \log_2(2^x+2)>0, logo
\left|\log_2(2^x+2)\right|=\log_2(2^x+2).
Se x\ge 0:
\log_2(2^x+2)>2x
\iff 2^x+2>2^{2x}.
Tomando y=2^x\ge1, isso equivale a
y^2-y-2<0
\iff (y-2)(y+1)<0.
Assim, 1\le y<2, ou seja, 0\le x<1.
Se x<0:
\log_2(2^x+2)>-2x
\iff 2^x+2>2^{-2x}.
Tomando y=2^x\in(0,1) e multiplicando por y^2>0, obtemos
y^3+2y^2-1>0.
A função f(y)=y^3+2y^2-1 é crescente para y>0, e
f\left(\frac{\sqrt5-1}{2}\right)=0.
Portanto, y>\frac{\sqrt5-1}{2}, o que implica
x>\log_2\left(\frac{\sqrt5-1}{2}\right).
Com x<0, a solução neste caso é
\log_2\left(\frac{\sqrt5-1}{2}\right)<x<0.
Unindo os dois casos:
Resposta final: \displaystyle \log_2\left(\frac{\sqrt5-1}{2}\right)<x<1.
Respondeu · mesma resposta final de outras 8 IAs
x\in\left(\log_2\dfrac{\sqrt5-1}{2},\;1\right), isto é, -\log_2\dfrac{1+\sqrt5}{2}<x<1.
levou 17 s
Como pensou
Since 2^x+2 > 2 always, the absolute value simplifies to log₂(2^x+2) itself, so the inequality becomes 2^x + 2 > 4^|x|. I'm now splitting into cases based on the sign of x to handle the absolute value.
For x≥0, setting t=2^x leads to (t-2)(t+1)<0, giving 1≤t<2, so 0≤x<1. For x<0, substitution gives t³+2t²-1>0, which factors as (t+1)(t²+t-1)>0, yielding t>(√5-1)/2 and thus log₂((√5-1)/2)<x<0. Combining both cases, the solution is the interval (log₂((√5-1)/2), 1).
Resolução
Como 2^x+2>2 para todo x\in\mathbb{R}, temos \log_2(2^x+2)>1>0, logo |\log_2(2^x+2)|=\log_2(2^x+2).
A inequação fica:
\log_2(2^x+2)>|2x| \iff 2^x+2>2^{|2x|}=4^{|x|}.
Caso 1: x\ge 0. Seja t=2^x\ge 1. Então
t+2>t^2 \iff t^2-t-2<0 \iff (t-2)(t+1)<0 \iff -1<t<2.
Com
t\ge1:
1\le 2^x<2 \Rightarrow 0\le x<1.
Caso 2: x<0. Seja t=2^x\in(0,1). Então 4^{|x|}=4^{-x}=\dfrac{1}{t^2} e
t+2>\frac{1}{t^2}\iff t^3+2t^2-1>0 \iff (t+1)(t^2+t-1)>0.
Como
t+1>0, precisamos de
t^2+t-1>0, isto é,
t>\dfrac{\sqrt5-1}{2} (a raiz negativa é descartada). Com
t<1:
\frac{\sqrt5-1}{2}<2^x<1 \Rightarrow \log_2\!\left(\frac{\sqrt5-1}{2}\right)<x<0.
Juntando os casos (o ponto x=0 está incluído, pois \log_2 3>0):
\log_2\!\left(\frac{\sqrt5-1}{2}\right)<x<1,
ou, equivalentemente,
-\log_2\!\left(\dfrac{1+\sqrt5}{2}\right)<x<1.
Resposta final: x\in\left(\log_2\dfrac{\sqrt5-1}{2},\;1\right), isto é, -\log_2\dfrac{1+\sqrt5}{2}<x<1.
Respondeu · mesma resposta final de outras 8 IAs
\log_2\!\left(\frac{\sqrt5-1}{2}\right) < x < 1
levou 12 s
Como pensou
Setting t = 2^x to simplify, since log2(t+2) is always positive, the equation becomes comparing log2(t+2) with |log2(t²)|. For x ≥ 0, I need log2(t+2) > log2(t²), which reduces to t² - t - 2 < 0, factoring to find where (t-2)(t+1) < 0.
For x < 0, I set up log2(t+2) > -2x = log2(1/t²), leading to t³ + 2t² - 1 > 0, which factors as (t+1)(t²+t-1) > 0. Since t > 0, I need t² + t - 1 > 0, giving t > (√5-1)/2, which translates to x > log2((√5-1)/2).
Combining both cases, the solution is log2((√5-1)/2) < x < 1.
Substituição. Seja t = 2^x > 0.
Como t + 2 > 2, temos \log_2(t+2) > 1 > 0. Logo o lado esquerdo é simplesmente \log_2(2^x+2).
Além disso, |2x| = |\log_2(t^2)|.
Caso 1: x \ge 0, ou seja, t \ge 1.
Aqui |2x| = 2x = \log_2 t^2. A inequação fica:
\log_2(t+2) > \log_2 t^2 \iff t+2 > t^2 \iff t^2 - t - 2 < 0 \iff (t-2)(t+1) < 0.
Como t+1 > 0, isso equivale a t < 2, isto é, x < 1.
Solução neste caso: 0 \le x < 1.
Caso 2: x < 0, ou seja, 0 < t < 1.
Aqui |2x| = -2x = \log_2 \dfrac{1}{t^2}. A inequação fica:
\log_2(t+2) > \log_2\frac{1}{t^2} \iff t^2(t+2) > 1 \iff t^3 + 2t^2 - 1 > 0.
Como t = -1 é raiz do polinômio, ele se fatora:
t^3 + 2t^2 - 1 = (t+1)(t^2 + t - 1).
Como t + 1 > 0, a condição equivale a t^2 + t - 1 > 0. Para t > 0, isso significa
t > \frac{\sqrt5 - 1}{2}.
Esse valor é menor que 1, então é compatível com t < 1. Voltando para x:
\log_2\frac{\sqrt5-1}{2} < x < 0.
Conclusão. Unindo os dois casos:
\log_2\left(\frac{\sqrt5-1}{2}\right) < x < 1.
Resposta final: \log_2\!\left(\frac{\sqrt5-1}{2}\right) < x < 1
Respondeu · mesma resposta final de outras 8 IAs
x ∈ ( log₂((√5 − 1)/2) , 1 )
levou 17 s
Como pensou
Setting t=2^x, I note log2(t+2) is always greater than 1 since t+2>2. For x≥0, the inequality log2(2^x+2) > 2x simplifies by substitution into a quadratic in t, giving t²-t-2<0.
Factoring gives (t-2)(t+1)<0, so t<2, meaning x<1, combined with x≥0 gives 0≤x<1. For x<0, the inequality becomes log2(2^x+2) > -2x, leading to t³+2t²-1>0, which factors as (t+1)(t²+t-1)>0, so I need t²+t-1>0, giving t > (√5-1)/2, hence x > log2((√5-1)/2). Combining both cases, the solution is log2((√5-1)/2) < x < 1.
Simplificação. Para todo x real, 2ˣ + 2 > 2, logo log₂(2ˣ + 2) > 1 > 0. Então |log₂(2ˣ + 2)| = log₂(2ˣ + 2), e a inequação fica
log₂(2ˣ + 2) > |2x|.
Seja t = 2ˣ > 0. Como log₂ é crescente, analisamos dois casos.
Caso 1: x ≥ 0. Aqui |2x| = 2x, então
log₂(2ˣ + 2) > 2x ⇔ 2ˣ + 2 > 2²ˣ ⇔ t² − t − 2 < 0 ⇔ (t − 2)(t + 1) < 0.
Como t + 1 > 0, isso equivale a t < 2, ou seja, x < 1. Logo 0 ≤ x < 1.
Caso 2: x < 0. Aqui |2x| = −2x, então
log₂(2ˣ + 2) > −2x ⇔ 2ˣ + 2 > 2⁻²ˣ ⇔ t + 2 > 1/t².
Multiplicando por t² > 0, obtemos t³ + 2t² − 1 > 0. Fatorando, t³ + 2t² − 1 = (t + 1)(t² + t − 1).
Como t + 1 > 0, a condição vira t² + t − 1 > 0. Para t > 0, isso dá t > (√5 − 1)/2, ou seja, x > log₂((√5 − 1)/2).
Como (√5 − 1)/2 ≈ 0,618 < 1, esse valor de x é negativo. Logo log₂((√5 − 1)/2) < x < 0.
União dos casos:
log₂((√5 − 1)/2) < x < 1.
Resposta final: x ∈ ( log₂((√5 − 1)/2) , 1 )
Respondeu · mesma resposta final de outras 8 IAs
-\log_2\bigl((1+\sqrt{5})/2\bigr) < x < 1
levou 1 min 27 s
Como pensou
The task is to find all real values of x satisfying |log₂(2^x + 2)| > |2x|.
The domain is all real numbers. I factored the log argument as 2(2^{x-1} + 1) and rewrote it as 1 + log₂(2^{x-1} + 1), but this doesn't immediately help solve the inequality.
I noted that log₂(2^x + 2) is always positive since 2^x + 2 > 2. I also considered squaring both sides of the inequality, but that approach seems difficult to solve directly.
The absolute value can be dropped because the log is always positive. The inequality simplifies to log₂(2^x + 2) > 2|x|.
I converted the inequality to 2^x + 2 > 2^{2|x|}. I then substituted y = 2^x and solved for x ≥ 0, finding 0 ≤ x < 1.
I verified the boundary values for x ≥ 0. Now I am solving the case x < 0 by substituting t = −x and z = 2^t, which transforms the inequality into 1/z + 2 > z².
I multiplied both sides of the inequality by z to get z³ − 2z − 1 < 0. I checked g(1) = −2 and g(2) = 3, confirming a root between 1 and 2. I also found that z = −1 is a root, so z + 1 divides the cubic polynomial.
I completed the polynomial division and found that z² − z − 1 = 0 has roots at (1 ± √5)/2. Since z > 1, the factor z² − z − 1 is negative exactly when 1 < z < (1+√5)/2. This gives the upper bound for the case x < 0.
I now have the full solution. For x < 0, the inequality holds when −log₂((1+√5)/2) < x < 0. Combining this with the earlier result for x ≥ 0, the complete set of solutions is −log₂((1+√5)/2) < x < 1.
The lower bound can be simplified to log₂((√5 − 1)/2), which equals log₂(1/φ). This is the exact value of the negative root.
Resolução
O domínio da inequação é \mathbb{R}, pois 2^x + 2 > 0 para todo x \in \mathbb{R}.
Observe que 2^x + 2 > 2 para todo x \in \mathbb{R}, de modo que
\log_2(2^x + 2) > 1 > 0.
Assim, |\log_2(2^x + 2)| = \log_2(2^x + 2), e a inequação se reduz a
\log_2(2^x + 2) > 2|x|,
ou, equivalentemente (já que a função \log_2 é crescente),
2^x + 2 > 2^{2|x|}.
Caso x \geq 0. A inequação torna-se 2^x + 2 > 2^{2x}. Substituindo y = 2^x (donde y \geq 1), obtém-se y + 2 > y^2, isto é,
y^2 - y - 2 < 0 \iff (y - 2)(y + 1) < 0.
Como y + 1 > 0, segue que y < 2. Combinando com y \geq 1, tem-se 1 \leq y < 2, ou seja, 0 \leq x < 1.
Caso x < 0. A inequação torna-se 2^x + 2 > 2^{-2x}. Substituindo t = -x > 0 e z = 2^t > 1, obtém-se 2 + 1/z > z^2, ou ainda (multiplicando por z > 0)
z^3 - 2z - 1 < 0.
Fatorando, z^3 - 2z - 1 = (z + 1)(z^2 - z - 1). Como z > 1, tem-se z + 1 > 0, e a inequação se reduz a z^2 - z - 1 < 0. As raízes de z^2 - z - 1 = 0 são (1 \pm \sqrt{5})/2, de modo que
z^2 - z - 1 < 0 \iff \frac{1 - \sqrt{5}}{2} < z < \frac{1 + \sqrt{5}}{2}.
Intersectando com z > 1, conclui-se que 1 < z < (1 + \sqrt{5})/2. Portanto,
0 < t < \log_2\Bigl(\frac{1 + \sqrt{5}}{2}\Bigr),
isto é,
-\log_2\Bigl(\frac{1 + \sqrt{5}}{2}\Bigr) < x < 0.
(Equivalentemente, o extremo inferior pode ser reescrito como \log_2\bigl((\sqrt{5} - 1)/2\bigr).)
Unindo os dois casos, a solução é o intervalo aberto
-\log_2\Bigl(\frac{1 + \sqrt{5}}{2}\Bigr) < x < 1.
(Nos extremos há igualdade, que é excluída pela inequação estrita; fora do intervalo a desigualdade se inverte, como se verifica diretamente pela análise de casos.)
Resposta final: -\log_2\bigl((1+\sqrt{5})/2\bigr) < x < 1
Respondeu · mesma resposta final de outras 8 IAs
\displaystyle \left(\log_2\!\left(\frac{\sqrt5-1}{2}\right),\,1\right)
levou 20 s
Como pensou
We need answer in Portuguese, solve inequality. Need think thoroughly. Problem: Determine all real x such that | log_2(2^x + 2) | > | 2x |.
We need solve. Let's analyze.
Inequality: | log2(2^x + 2) | > |2x| = 2|x|? Actually |2x| = 2|x|. Since domain 2^x +2 >0 always.
Let t = 2^x >0. Then x = log2 t. Inequality | log2(t+2) | > | 2 log2 t | = 2 | log2 t |, t>0. Or maybe use base 2 logs. Let y = log2(t+2). Then x = log2 t. Relationship y = log2(2^x+2) = log2(2(2^{x-1}+1))? Maybe.
Inequality | log2(2^x+2) | > 2 |x|. Let f(x) = log2(2^x+2). Need absolute.
We can consider x >=0 and x<0 separately.
First note f(x)>? For all x, 2^x+2>1 so log>0? Wait log2(2^x+2): since 2^x+2 >1? 2^x >0, so +2 >2 >1, log2 >1 actually >1. So f(x)=log2(2^x+2) >1 for all x. Thus |f(x)| = f(x). So inequality f(x) > 2|x|.
So solve log2(2^x + 2) > 2|x|.
For x >=0: inequality log2(2^x+2) > 2x. Need solve. Let's manipulate.
log2(2^x+2) > 2x = log2(2^{2x})? Since 2x = log2(2^{2x})? Actually 2^{2x} = (2^x)^2. Yes log2(4^x)? Wait 2^{2x} = 4^x, log2(2^{2x})=2x. So inequality log2(2^x+2) > log2(2^{2x}) equivalent (base 2 increasing) 2^x+2 > 2^{2x}. Let u = 2^x >1 for x>=0 (u>=1). Inequality u + 2 > u^2 => u^2 - u - 2 <0 => (u-2)(u+1)<0 => -1<u<2. Since u>0, u<2. Also u>=1 for x>=0. So 1 <= u <2 => 0 <= x <1. Since u=2^x <2 => x <1. Equality? Strict > so u<2 => x<1. At x=0? f(0)=log2(3)=1.585 >0 yes; inequality 1.585>0 true. So [0,1).
For x <0: |x|=-x. Need f(x) > -2x = log2(2^{-2x})? Since -2x positive. Inequality log2(2^x+2) > -2x = log2(2^{-2x})? Actually 2^{-2x} = (2^x)^{-2}? Since x<0, -2x>0. log2(2^{-2x}) = -2x. So inequality equivalent 2^x+2 > 2^{-2x}. Let u = 2^x. For x<0, 0<u<1. Inequality u+2 > u^{-2} = 1/u^2. Multiply positive u^2: u^3 + 2u^2 >1. Need solve for u in (0,1): u^3 + 2u^2 -1 >0. Let g(u)=u^3+2u^2-1. Find roots. g(0)=-1, g(1)=2>0. Derivative 3u^2+4u = u(3u+4)>0 for u>0, so g increasing on (0,1). Thus unique root r in (0,1). Then x < log2 r (since u<r => x< log2 r, because log2 increasing; wait g>0 => u>r, so 2^x > r => x > log2 r? Since r<1, log2 r negative. If x<0, u=2^x, u>r means x > log2 r? Let's check: 2^x increasing; if u>r, x > log2 r (which negative). Since r in (0,1), log2 r negative. So x in (log2 r, 0). Need combine? Also maybe x=0? x=0 not x<0.
Need determine root r exactly? g(u)=u^3+2u^2-1=0. Does it factor? Try u=-1: -1+2-1=0 yes! u=-1 root. Factor (u+1)(u^2+u-1)? Let's divide: u^3+2u^2+0u-1 divided by u+1 = u^2 + u -1? (u+1)(u^2+u-1)=u^3+u^2-u + u^2+u-1 = u^3+2u^2+0u -1 yes. Roots: u=-1, u = [-1 ± sqrt(1+4)]/2 = (-1 ± sqrt5)/2. Positive root (-1+√5)/2 ≈0.618. Negative root <0. So r=(√5 -1)/2. Nice. Thus u>r => x > log2((√5-1)/2). Since log2 r = log2((√5-1)/2). Negative.
So solution x in ( log2((√5-1)/2), 0 ) ∪ [0,1)? Wait intervals: For x<0: x > log2 r. Since log2 r negative. So (log2 r, 0). For x>=0: x <1 and x>=0 => [0,1). Combine: ( log2((√5-1)/2), 1 ). But note x=0 included from nonnegative interval; x<0 interval (log2 r,0) open at 0, so union with [0,1) is (log2 r,1). Need check at x=0 inequality strict yes. At x? x=log2 r? For x=log2 r, equality? f(x)=log2(r+2); -2x = log2(r^{-2})? Since x=log2 r => 2^x = r. Then inequality f(x)> -2x equivalent r+2 > r^{-2} -> r^3+2r^2? Wait equality when g(r)=0 => r^3+2r^2=1 -> r+2 = r^{-2}? Multiply r^2: r^3 + 2r^2 =1 yes. So equality, not included. At x=1? f(1)=log2(4)=2; |2x|=2, equality not >. So open at 1. At x=0 strict.
But need ensure for x> ? If x>1, inequality false? Check x=2: f(2)=log2(6)=2.585, 2x=4, false. Yes.
Thus answer: (log_2((√5-1)/2), 1). But need verify with alternative derivation perhaps.
Let's solve carefully.
Inequality: |log2(2^x+2)| > |2x| = 2|x|. Since 2^x+2 >1 -> log2(...) >1>0, so absolute unnecessary first. So inequality becomes log2(2^x+2) > 2|x|.
Define h(x)=log2(2^x+2)-2|x|. Need h(x)>0.
Because absolute in RHS, split x >=0 and x<0.
Case x >=0: h = log2(2^x+2) - 2x. Exponentiate: equivalent 2^x+2 > 2^{2x} = 4^x? Wait 2^{2x} = (2^x)^2. Yes. Let y=2^x. y>=1. Inequality y+2 > y^2 -> y^2 - y -2 <0 -> (y-2)(y+1)<0 -> y<2 because y>0. So y in [1,2): x in [0,1). Good.
Case x <0: h = log2(2^x+2) + 2x? Actually -2|x|=-2(-x)=+2x, so inequality log2(2^x+2) > -2x. Since -2x positive? x<0, -2x positive yes. Exponentiate: 2^x+2 > 2^{-2x} = (2^x)^{-2}. Let y=2^x, 0<y<1. Inequality y+2 > y^{-2}. Multiply y^2: y^3+2y^2 >1. Let P(y)=y^3+2y^2-1. P(0)=-1, P(1)=2, strictly increasing? derivative 3y^2+4y >0 for y>0, yes. Root positive r =? P(r)=0. Factor: r^3+2r^2-1=(r+1)(r^2+r-1). Positive root (-1+√5)/2. Thus P(y)>0 iff y>r. Since y=2^x increasing, x>log2 r = log2((√5-1)/2). Also x<0. So x in (log2 r, 0). Combine: (log2 r, 0) ∪ [0,1) = (log2 r,1). Since log2 r <0.
Check endpoint x=0: Is it included? Our nonnegative case includes 0. From negative case upper open at 0, but union includes. Inequality at x=0: log2(3) ≈1.585 >0 true. Yes.
At x=1: log2(4)=2 >2? No equality. Not included.
At x=log2 r: equality? Let's verify: x=log2 r, 2^x=r. RHS -2x? Since x<0, |2x|=-2x = -2 log2 r = log2(r^{-2})? Because -2 log2 r = log2(r^{-2}). LHS log2(r+2). Equality iff r+2 = r^{-2} -> r^3+2r^2=1, true by P(r)=0. So equality at lower endpoint; open.
Thus final \(x \in \left(\log_2\frac{\sqrt5-1}{2}, 1\right)\). Need maybe express as \(\log_2(\sqrt5-1)-1\)? Since log2((√5-1)/2) = log2(√5-1) -1. Could use.
Need produce solution in Portuguese, item by item maybe no items. The question has "Determine todos os valores x ∈ R tais que ...". Need final answer. We should show development. The prompt says "Resolva a questão como um candidato faria na prova: mostre o desenvolvimento em português, item por item quando a questão tiver itens (a, b, c…). Termine..." So provide step-by-step. Need perhaps mention convention natural, etc. The image has convenções but not needed except R. We should use notation.
Let's write clean solution.
Potential pitfalls: Domain: 2^x + 2 >0 always. Absolute of log: Since 2^x+2>1, log2(2^x+2)>0? Wait log2(y)>0 iff y>1. y=2^x+2 >2>1, yes >0. So | log2(...) | = log2(...). Good.
Then solve:
\[
\log_2(2^x+2)>2|x|.
\]
Separate.
1) x ≥0:
\[
\log_2(2^x+2)>2x=\log_2(2^{2x})
\]
since log2 strictly increasing:
\[
2^x+2>2^{2x} \iff (2^x)^2 - 2^x -2 <0.
\]
Set t=2^x ≥1:
\[
t^2-t-2<0 \iff (t-2)(t+1)<0 \iff -1<t<2.
\]
With t≥1: \(1≤t<2\), so \(0≤x<1\).
2) x<0:
RHS |2x| = -2x = log2(2^{-2x})? Let's check: log2(2^{-2x}) = -2x yes. So inequality:
\[
\log_2(2^x+2) > -2x = \log_2(2^{-2x})
\]
so
\[
2^x+2 > 2^{-2x}.
\]
Set t=2^x, 0<t<1:
\[
t+2 > t^{-2}
\]
multiply t^2>0:
\[
t^3+2t^2 >1
\]
\[
t^3+2t^2 -1 >0
\]
Factor:
\[
(t+1)(t^2+t-1)>0
\]
Since t+1>0, need \(t^2+t-1>0\). For t in (0,1), roots of t^2+t-1 are \((-1±√5)/2\). Positive root \(r=(-1+√5)/2\). Quadratic >0 for t > r (since leading positive and t positive; indeed for t>r). Thus \(t > r\), so \(2^x > r\). Since r<1, log2 r <0, and \(x > \log_2 r\). Also x<0.
Need be precise: t^2+t-1>0 for \(t < (-1-√5)/2\) or \(t > (-1+√5)/2\). Since t>0, t>r. So \(2^x > r\) => \(x > \log_2 r\) because r<1? Let's check: 2^x increasing; if r<1 and 2^x>r, since log2 r negative, yes x > log2 r. Eg r=0.618, x > -0.693? log2(0.618)=-0.693. If x=-0.5, 2^-0.5=0.707>0.618 yes. If x=-1, 0.5<0.618 no. So yes.
So case 2 solution: \( \log_2 r < x <0\).
Combine: \(x \in (\log_2 r, 0) \cup [0,1) = (\log_2 r,1)\).
Thus final.
Need maybe include "note that log2 r = log2((√5-1)/2) = log2(√5-1)-1". Could put final.
Let's craft final answer in Portuguese. The prompt wants final line "Resposta final: ...". We can include development before.
Potential issue: In case x<0, when exponentiating, \(2^{-2x}\). Since x<0, -2x>0, \(2^{-2x}\) >1? Wait if x=-0.5, 2^{-2(-0.5)}=2^1=2; yes. Inequality t+2 > t^{-2}. t<1, t^{-2}>1. Okay.
Let's ensure no missed cases: x=0 included in first case. x=1? t=2, t^2 -t -2 =4-2-2=0, inequality false (strict < for t^2-t-2<0? At t=2 equals 0; not <0). So x<1. Good.
At lower endpoint equality, open.
Final interval:
\[
\left(\log_2\frac{\sqrt5-1}{2},\,1\right).
\]
Equivalent \(\left(\log_2(\sqrt5-1)-1,\,1\right).
Need maybe use \(\log_2\) notation in Portuguese.
Let's write final solution elegantly.
Seja x\in\mathbb R. Como 2^x+2>2>1, temos
\log_2(2^x+2)>0,
logo
\left|\log_2(2^x+2)\right|=\log_2(2^x+2).
Assim, a inequação equivale a
\log_2(2^x+2)>2|x|.
Separamos em dois casos.
1) x\ge 0
Neste caso, |2x|=2x. Logo
\log_2(2^x+2)>2x=\log_2(2^{2x}),
e como a função \log_2 é crescente,
2^x+2>2^{2x}.
Fazendo t=2^x\ge 1, obtemos
t+2>t^2 \iff t^2-t-2<0 \iff (t-2)(t+1)<0.
Como t\ge1, segue 1\le t<2, isto é,
0\le x<1.
2) x<0
Neste caso, |2x|=-2x=\log_2(2^{-2x}). Assim,
\log_2(2^x+2)>-2x=\log_2(2^{-2x}),
donde
2^x+2>2^{-2x}.
Fazendo t=2^x, com 0<t<1, temos
t+2>t^{-2}.
Multiplicando por t^2>0:
t^3+2t^2>1 \iff t^3+2t^2-1>0.
Fatorando:
t^3+2t^2-1=(t+1)(t^2+t-1).
Como t+1>0, devemos ter
t^2+t-1>0.
As raízes de t^2+t-1=0 são
t=\frac{-1\pm\sqrt5}{2}.
Como 0<t<1, a raiz positiva relevante é
t=\frac{\sqrt5-1}{2}.
Portanto,
2^x>\frac{\sqrt5-1}{2}.
Logo
x>\log_2\!\left(\frac{\sqrt5-1}{2}\right).
Juntando com x<0, obtemos
\log_2\!\left(\frac{\sqrt5-1}{2}\right)<x<0.
União dos casos
De 1), temos 0\le x<1.
De 2), temos \log_2\!\left(\frac{\sqrt5-1}{2}\right)<x<0.
Portanto,
\log_2\!\left(\frac{\sqrt5-1}{2}\right)<x<1.
Resposta final: \displaystyle \left(\log_2\!\left(\frac{\sqrt5-1}{2}\right),\,1\right)
Respondeu · mesma resposta final de outras 8 IAs
\log_2\left(\frac{\sqrt{5} - 1}{2}\right) < x < 1
levou 26 s
Como pensou
**Exploring Inequality Domain**
I'm currently focused on defining the operational scope of this inequality. My initial thought is to confirm the domain where $\log_2(2^x + 2)$ is well-defined. This seems straightforward, as $2^x + 2$ is always positive for any real $x$.
**Confirming Domain and Simplifying**
I've confirmed that $\log_2(2^x + 2)$ is always positive for any real $x$. This allows me to remove the absolute value directly, simplifying the inequality to $\log_2(2^x + 2) > |2x|$.
**Solving Inequality Branches**
I am now analyzing the two branches of the inequality. For the case $2x \ge 0$, I'm simplifying the expression by removing the absolute value and solving the resulting inequality $2^x + 2 > 2^{2x}$. The second branch, $2x < 0$, requires careful handling of the exponentiation after introducing the term $y = 2^x$, leading to the polynomial $y^3 + 2y^2 - 1 > 0$.
**Solving Inequality Branches**
I am now analyzing the two branches of the inequality. For the case $2x \ge 0$, I'm simplifying the expression by removing the absolute value and solving the resulting inequality $2^x + 2 > 2^{2x}$. The second branch, $2x < 0$, requires careful handling of the exponentiation after introducing the term $y = 2^x$, leading to the polynomial $y^3 + 2y^2 - 1 > 0$.
**Analyzing Polynomial Roots**
I've established that for the $2x < 0$ case, the relevant polynomial is $y^3 + 2y^2 - 1 > 0$, where $y = 2^x$. I'm now focusing on its roots to determine the intervals where it holds true.
**Determining Solution Intervals**
I've found that for $y > 0$, the condition $y^2 + y - 1 > 0$ is satisfied when $y > \frac{\sqrt{5} - 1}{2}$. Translating this back to $x$, this means $x > \log_2\left(\frac{\sqrt{5} - 1}{2}\right)$. Combining this with the condition $x < 1$ from the $2x \ge 0$ case yields the solution interval.
**Finalizing Solution Range**
The analysis confirms a solution range for $x$ from $\log_2\left(\frac{\sqrt{5} - 1}{2}\right)$ up to, but not including, 1. This interval satisfies both branches of the original inequality.
**Confirming Solution Intervals**
My verification of the derived solution intervals through direct testing of values, including boundary conditions and points within and outside the expected ranges, has yielded results perfectly consistent with the established inequality. This reinforces confidence in the solution obtained.
**Resolving Polynomial Roots**
I've confirmed that for $y > 0$, the condition $y^2 + y - 1 > 0$ holds when $y > \frac{\sqrt{5}-1}{2}$. This translates back to $x$ as $x > \log_2\left(\frac{\sqrt{5}-1}{2}\right)$.
Para resolver a inequação dada, analisamos inicialmente a expressão no interior do primeiro módulo.
Para todo x \in \mathbb{R}, temos 2^x > 0, o que implica:
2^x + 2 > 2
Como a função logarítmica na base 2 é estritamente crescente (base 2 > 1):
\log_2(2^x + 2) > \log_2(2) = 1 > 0
Logo, \log_2(2^x + 2) é estritamente positivo para todo x \in \mathbb{R}, permitindo eliminar o módulo:
|\log_2(2^x + 2)| = \log_2(2^x + 2)
A inequação original reescreve-se como:
\log_2(2^x + 2) > |2x|
Pela definição de valor absoluto, para uma grandeza positiva no lado esquerdo, essa inequação é equivalente ao sistema de inequações simultâneas:
-\log_2(2^x + 2) < 2x < \log_2(2^x + 2)
Vamos resolver cada desigualdade separadamente:
1) Primeira desigualdade: 2x < \log_2(2^x + 2)
Como a função exponencial f(t) = 2^t é estritamente crescente:
2^{2x} < 2^x + 2
Fazendo a mudança de variável y = 2^x (com y > 0):
y^2 - y - 2 < 0
Fatorando o trinômio do segundo grau:
(y - 2)(y + 1) < 0 \implies -1 < y < 2
Como y = 2^x > 0, temos:
0 < 2^x < 2 \implies 2^x < 2^1 \implies x < 1
2) Segunda desigualdade: 2x > -\log_2(2^x + 2)
Multiplicando por -1 e aplicando a função exponencial de base 2:
\log_2(2^x + 2) > -2x \iff 2^x + 2 > 2^{-2x}
Em termos de y = 2^x (com y > 0):
y + 2 > \frac{1}{y^2}
Multiplicando ambos os lados por y^2 > 0:
y^3 + 2y^2 - 1 > 0
Observando que y = -1 é raiz de y^3 + 2y^2 - 1 = 0, podemos fatorar o polinômio:
(y + 1)(y^2 + y - 1) > 0
Como y > 0, o fator (y + 1) é estritamente positivo (y + 1 > 1 > 0). Portanto, a desigualdade reduz-se a:
y^2 + y - 1 > 0
As raízes da equação y^2 + y - 1 = 0 são dadas por:
y = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}
Como y > 0, a solução para y é:
y > \frac{\sqrt{5} - 1}{2}
Retornando para a variável x:
2^x > \frac{\sqrt{5} - 1}{2} \iff x > \log_2\left(\frac{\sqrt{5} - 1}{2}\right) = \log_2(\sqrt{5} - 1) - 1
Conclusão
Fazendo a interseção dos intervalos obtidos em (1) e (2):
\log_2\left(\frac{\sqrt{5} - 1}{2}\right) < x < 1
Resposta final: \log_2\left(\frac{\sqrt{5} - 1}{2}\right) < x < 1
Respondeu · mesma resposta final de outras 8 IAs
levou 40 s
Como pensou
**Analyzing the Inequality**
I am currently working through the mathematical inequality $| \log_2(2^x + 2) | > |2x|$. My immediate focus is on understanding the behavior of the expression $2^x + 2$ within the logarithm.
**Simplifying the Inequality**
I've determined that $\log_2(2^x + 2)$ is always positive, allowing me to remove one absolute value. The inequality now simplifies to $\log_2(2^x + 2) > |2x|$. I'm currently exploring the case where $x \ge 0$.
**Solving for Negative Exponents**
I've successfully analyzed the case where $x < 0$. Substituting $t=2^x$ transformed the inequality into $(t+1)(t^2+t-1) > 0$. Since $t>0$ for $x<0$, I focused on $t^2+t-1 > 0$. The roots of the quadratic are $\frac{-1 \pm \sqrt{5}}{2}$.
**Refining the Solution**
I've successfully analyzed the negative case, reducing it to a quadratic inequality in terms of $t=2^x$. This led to a specific range for $x$ between $\log_2\left(\frac{\sqrt{5} - 1}{2}\right)$ and 0.
Para determinar todos os valores reais de x que satisfazem a inequação |\log_2(2^x + 2)| > |2x|, observemos inicialmente o comportamento do logaritmando.
Para todo x \in \mathbb{R}, temos 2^x > 0, o que implica:
2^x + 2 > 2
Como a função logarítmica de base 2 é estritamente crescente, segue que:
\log_2(2^x + 2) > \log_2(2) = 1 > 0
Portanto, a expressão dentro do módulo no membro esquerdo é sempre positiva, permitindo reescrever a inequação original simplesmente como:
\log_2(2^x + 2) > |2x|
Dividimos a análise em dois casos, de acordo com o sinal de x:
Caso 1: x \ge 0
Neste caso, |2x| = 2x, e a inequação torna-se:
\log_2(2^x + 2) > 2x \iff 2^x + 2 > 2^{2x}
Fazendo a substituição t = 2^x, como x \ge 0, temos t \ge 1. A inequação em t fica:
t + 2 > t^2 \iff t^2 - t - 2 < 0 \iff (t - 2)(t + 1) < 0
As raízes da equação associada são t = -1 e t = 2, logo a solução da inequação quadrática é -1 < t < 2. Intersectando com a condição t \ge 1, obtemos:
1 \le t < 2 \iff 1 \le 2^x < 2 \iff 0 \le x < 1
Caso 2: x < 0
Neste caso, |2x| = -2x, e a inequação torna-se:
\log_2(2^x + 2) > -2x \iff 2^x + 2 > 2^{-2x}
Utilizando novamente a substituição t = 2^x, a condição x < 0 implica 0 < t < 1. A inequação em t assume a forma:
t + 2 > \frac{1}{t^2}
Como t^2 > 0, podemos multiplicar ambos os lados por t^2 mantendo o sentido da desigualdade:
t^3 + 2t^2 - 1 > 0
Notando que t = -1 é raiz do polinômio P(t) = t^3 + 2t^2 - 1, podemos fatorá-lo como:
t^3 + 2t^2 - 1 = (t + 1)(t^2 + t - 1) > 0
Como 0 < t < 1, o fator (t + 1) é estritamente positivo. Assim, a desigualdade equivale a:
t^2 + t - 1 > 0
As raízes de t^2 + t - 1 = 0 são dadas por:
t = \frac{-1 \pm \sqrt{5}}{2}
Para que a expressão quadrática seja positiva, devemos ter t < \frac{-1 - \sqrt{5}}{2} ou t > \frac{\sqrt{5} - 1}{2}. Intersectando esse resultado com o intervalo 0 < t < 1, obtemos:
\frac{\sqrt{5} - 1}{2} < t < 1
Voltando para a variável x:
\frac{\sqrt{5} - 1}{2} < 2^x < 1 \iff \log_2\left(\frac{\sqrt{5} - 1}{2}\right) < x < 0
Conclusão:
Unindo as soluções encontradas no Caso 1 e no Caso 2, obtemos o conjunto solução para x \in \mathbb{R}:
\log_2\left(\frac{\sqrt{5} - 1}{2}\right) < x < 1 \quad \text{ou} \quad \log_2(\sqrt{5} - 1) - 1 < x < 1
Resposta final: ]log₂(√5 - 1) - 1, 1[
