0
$\begingroup$

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 $log(b)$

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

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

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

$=$ $2*log(3)$

$=$ $2*1 = 2$

$\endgroup$
  • 1
    $\begingroup$ In the future, use $\log$ instead of $log$. $\endgroup$ – angryavian Sep 15 '14 at 19:55
0
$\begingroup$

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: $\sqrt{2}=\log_3 (x) \Rightarrow x= 3^{\sqrt{2}}$

$\endgroup$
1
$\begingroup$

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

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

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

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

$$\log_b2+\log_bb=3$$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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