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$$(log_{2}(x+1))^2=4$$ $$log_{2}(x+1)*log_{2}(x+1)=log_{2}16$$ $$x^{2}+2x-15=0$$ $$(x+1)*(x+1)=16$$ $$x^{2}+2x+1=16$$ $$x^{2}+2x-15=0$$ $$(x+5)(x-3)=0$$ $$x_1=-5; x_2=3$$ The solution is only $x_1=3$. Is this correct?

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marked as duplicate by Rick Decker, Stefan4024, Old John, Vedran Šego, azimut Dec 7 '13 at 20:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Why isn't my way of solving this correct? –  user114141 Dec 7 '13 at 18:30
    
Because $\log(xy)\neq\log(x)\log(y)$ –  user91500 Dec 7 '13 at 18:31
    
$$(logx)^2=logx\cdot\ logx$$ says in my book for maths. –  user114141 Dec 7 '13 at 18:33
    
$\log x\log x=(\log x)^2\neq\log(x^2)$ –  user91500 Dec 7 '13 at 18:34
    
@Algean: I know that's not correct, I used $(logx)^2=logx\cdot\ logx$, not $log(xy)=log(x)log(y)$. –  user114141 Dec 7 '13 at 18:36

4 Answers 4

up vote 3 down vote accepted

Your solution is incorrect. First note that $$\log_2\left(a^2 \right) \neq \left(\log_2(a) \right)^2$$ In your case, we have $$\left(\log_2(x+1) \right)^2 = 4 \implies \log_2(x+1) = \pm2 \implies x+1 = 2^{\pm2} \implies x+1 = 4 \text{ or }x+1= \dfrac14$$ Hence, $$x=3 \text{ or }x=-\dfrac34$$

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If $$(\log_2(x+1))^2=4$$ then $$\log_2(x+1)=\color{red}{+2}$$ or $$\log_2(x+1)=\color{blue}{-2}$$ so $x+1=2^{\color{red}{+2}}$ or $x+1=2^{\color{blue}{-2}}$.

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@SamiBenRomdhane: انشاء الله تبارک و تعالی :-) –  Babak S. Dec 19 '13 at 19:11

$$(\log_{2}(x+1))^2=4\iff\log_{2}(x+1)=\pm2$$ [1]$$\log_{2}(x+1)=2,x+1=2^2,x_1=3$$ [2]$$\log_{2}(x+1)=-2,x+1=2^{-2},x_2=-3/4$$

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Nice to see Adi. Miss u at FB. –  Babak S. Dec 7 '13 at 18:49

$$ (\log_2(x+1))^2 = 4 \Rightarrow \log_2(x+1) = 2 , - 2 $$

$$ \Rightarrow x + 1 = 2^2 , 2^{-2} \Rightarrow x = 3 \space \textrm{or} \space -\frac{3}{4} $$

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