Irvin Dukes

2021-12-23

How do you solve ${8}^{x}=4$?

Jack Maxson

Expert

${8}^{x}=4⇔x=\frac{2}{3}$
Explanation: ${8}^{x}=4$
Since $4=\left(2\right)\left(2\right)={2}^{2}$
we can say that
$8=4\left(2\right)=\left(2\right)\left(2\right)\left(2\right)=2\left({2}^{2}\right)={2}^{2+1}={2}^{3}$
So by changing notation we get
$⇔{\left({2}^{3}\right)}^{x}={2}^{2}$
and since ${\left({x}^{a}\right)}^{b}={x}^{ab}$, we can say
$⇔{2}^{3x}={2}^{2}$
Then we take the ${\mathrm{ln}}_{2}\left(x\right)$ of both sides
$⇔{\mathrm{ln}}_{2}\left({2}^{3x}\right)={\mathrm{ln}}_{2}\left({2}^{2}\right)$ This gives us
$⇔3x=2$
Then we divide both sides by 3 to isolate x
$⇔x=\frac{2}{3}$

ramirezhereva

Expert

Solve for x over the real numbers:
${8}^{x}=4$
Hint:
Eliminate the exponential from the left hand side.
Take the logarithm base 8 of both sides:
$x=\frac{2}{3}$

user_27qwe

Expert