# Solve Exponential equation $2^x=3$

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

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

• You are right and text book is wrong. – Bumblebee Jul 27 '18 at 13:41
• 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). – lulu Jul 27 '18 at 13:41
• 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? – Ethan Bolker Jul 27 '18 at 13:41
• change book! :P – gimusi Jul 27 '18 at 13:44
• 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! – Cesare Jul 27 '18 at 13:49

$$2^x=2^{(\log_2 3)}=3$$
$$\log_3 x=-2 \implies 3^{\log_3 x}=3^{-2} \implies x=\frac19$$