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I want to solve this exponential equation:

$$2^x=3$$

To do this, I apply the logarithm of base 2 to both sides of the equation:

$$\log_22^x = 3 \implies x\log_22 = \log_23 \implies x = \log_23$$

I would not be able to go on from here. My textbook suggests that the answer is $\frac 1 9$. How? I can't really wrap my head around this even though I know this is pretty easy. Any hints?

Edit: this is what the textbook adds:

$$\log_3x = -2$$

And then applies a function $3^t$ so that

$$x=3^{-2}= \frac 1 9$$

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    $\begingroup$ You are right and text book is wrong. $\endgroup$ – Bumblebee Jul 27 '18 at 13:41
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    $\begingroup$ You've gone as far as you can easily go. Analytic methods show that $\log_23\approx 1.584962501$. Certainly not $\frac 19$ (easy to see that, in fact, it is irrational). $\endgroup$ – lulu Jul 27 '18 at 13:41
  • $\begingroup$ Your answer is right and $1/9$ is wrong. The correct numerical value is about $1.58$. Are you sure you copied the problem correctly? $\endgroup$ – Ethan Bolker Jul 27 '18 at 13:41
  • $\begingroup$ change book! :P $\endgroup$ – gimusi Jul 27 '18 at 13:44
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    $\begingroup$ Ops, it seems that the book reports two distinct examples of logarithmic and exponential equations but it's not made clear from the text. Sorry and thanks everyone! $\endgroup$ – Cesare Jul 27 '18 at 13:49
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Yes your solution is absolutely correct, indeed by definition

$$2^x=2^{(\log_2 3)}=3$$

That seems the solution to

$$\log_3 x=-2 \implies 3^{\log_3 x}=3^{-2} \implies x=\frac19 $$

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