Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
    
Why isn't my way of solving this correct? – user114141 Dec 7 '13 at 18:30
    
Because $\log(xy)\neq\log(x)\log(y)$ – MATHEMATIKER Dec 7 '13 at 18:31
    
$$(logx)^2=logx\cdot\ logx$$ says in my book for maths. – user114141 Dec 7 '13 at 18:33
    
$\log x\log x=(\log x)^2\neq\log(x^2)$ – MATHEMATIKER 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
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, $$x=3 \text{ or }x=-\dfrac34$$

share|cite|improve this answer

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 $x+1=2^{\color{blue}{-2}}$.

share|cite|improve this answer
    
@SamiBenRomdhane: انشاء الله تبارک و تعالی :-) – S. Snape 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]$$\log_{2}(x+1)=-2,x+1=2^{-2},x_2=-3/4$$

share|cite|improve this answer
    
Nice to see Adi. Miss u at FB. – S. Snape Dec 7 '13 at 18:49

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

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

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.