Solve for $x$ for the following exponential equation $2^{2x+1} = 3^{2x+1}$. What am I doing wrong?

Question

$2^{2x+1} = 3^{2x+1}$

$2^1=3$?

Why can't I take $\log_2$ of both sides ?

Answer 1

You can but you're taking it incorrectly, it seems.

To solve this, rewrite it as

$(\frac{2}{3})^{2x+1}=1$.

Now it's clear that $2x+1=0$ which gives

$x=-\frac{1}{2}$.

Answer 2

$2^{2x+1}=3^{2x+1}$

When you take the $\log_2$ of both sides, the following happens:

$\log_2{2^{2x+1}}=\log_2{3^{2x+1}}$

${2x+1}=({2x+1})\log_2{3}$

And as you noticed, if $x≠-\frac{1}{2}$, then you get the inconsistent identity that

$1=\log_2{3}$, which isn't true. So what gives?

The fact, is, the problem was when we assumed that $x≠-\frac{1}{2}$. Division by zero is not well defined, and in fact it turns out the solution is when $x=-\frac{1}{2}$.

The reason for this lies in the question itself. Look at the graphs of $f=2^a$ and $f=3^a$. Naturally the $3^a$ grows faster than the $2^a$ graph. The only intersection point is when $a=0$.

In our problem, "$a=0$" means "$2x+1=0$" and consequently $x=-\frac{1}{2}$.

Answer 3

Yes you can, and you find: $$ 2x+1=(2x+1)\log_2 3 \iff (2x+1)(1-\log_2 3)=0 \iff 2x+1=0 \iff x=-\frac{1}{2} $$