I am trying to solve for $x$ in
$x^2=(16)^{2x}.$
So I started this way:
I took square root of both sides and got
$x=16^x$
Then I took the logarithm of both sides and got
$\log x=x \log 16.$
This is where I got stuck.
algebra-precalculusexponential functionexponentiation
I am trying to solve for $x$ in
$x^2=(16)^{2x}.$
So I started this way:
I took square root of both sides and got
$x=16^x$
Then I took the logarithm of both sides and got
$\log x=x \log 16.$
This is where I got stuck.
Best Answer
Contrary to some other answers,
$$x=-0.36424988978364795656$$ is the real solution. (Can be expressed in terms of the Lambert $W$ function; otherwise there is no analytical expression.)
https://www.wolframalpha.com/input/?i=x%5E2%3D256%5Ex