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If $b=log_3(x),$ what value of $x$ satisfies $log_b(log_3(x^2))=3?$

I just started learning this topic by myself. I wanted to know if my working is correct. If not can someone help me with this solution?

$log_b(\frac{(log(x^2)}{log(3)})$

$=$ $log_b(log(x^2))$

$=$ $log_b(2log(x))$

$=$ $\frac{2log(x)}{log(b)}$

Since $b=log_3(x)$, we can substitute that in for $log(b)$

$=$ $\frac{2log(x)}{log_3(x)}$

$=$ $2log(x)/\frac{log(x)}{log(3)}$

$=$ $2log(x)*\frac{log(3)}{log(x)}$

$=$ $2*log(3)$

$=$ $2*1 = 2$

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    $\begingroup$ In the future, use $\log$ instead of $log$. $\endgroup$
    – angryavian
    Commented Sep 15, 2014 at 19:55

2 Answers 2

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The question should be, I believe, what value of $\;b\;$ satisfies the given equality, as $\;x\;$ has no role given that $\;\log_3x=b\iff 3^b=x\;$:

$$\log_b\left(\log_3x^2\right)=3$$

$$\log_b\left(2\log_3x\right)=3$$

$$\log_b\left(2b\right)=3$$

$$\log_b2+\log_bb=3$$

$$\log_b2=2\iff b^2=2\iff \ldots$$

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Let's start with $\log_b (\log_3(x^2)) = 3$. It can be written as $\log_3(x^2)=b^3 \Rightarrow 2 \log_3 (x) = b^3 \Rightarrow 2b=b^3 \Rightarrow b(2-b^2)=0 \Rightarrow b=\sqrt{2}$ and not the negative root because $b$ has to be positive. A base cannot be negative. Thus we get: $\sqrt{2}=\log_3 (x) \Rightarrow x= 3^{\sqrt{2}}$

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