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

Question

21865 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-17T00:58:49

$8^{x} = 4 \Leftrightarrow x = \frac{2}{3}$ $8^x=4 <=> x=2/3$

$8^{x} = 4$ $8^x=4$

Since $4 = (2) (2) = 2^{2}$ $4=(2)(2)=2^2$

we can say that

$8 = 4 (2) = (2) (2) (2) = 2 (2^{2}) = 2^{2 + 1} = 2^{3}$ $8=4(2)=(2)(2)(2)=2(2^2)=2^(2+1)=2^3$

So by changing notation we get

$\Leftrightarrow {(2^{3})}^{x} = 2^{2}$ $<=> (2^3)^x=2^2$

and since ${(x^{a})}^{b} = x^{a b}$ $(x^a)^b=x^(ab)$ , we can say

$\Leftrightarrow 2^{3 x} = 2^{2}$ $<=> 2^(3x)=2^2$

Then we take the ${ln}_{2} (x)$ $ln_2(x)$ of both sides

$\Leftrightarrow {ln}_{2} (2^{3 x}) = {ln}_{2} (2^{2})$ $<=> ln_2(2^(3x))=ln_2(2^2)$

This gives us

$\Leftrightarrow 3 x = 2$ $<=> 3x=2$

Then we divide both sides by 3 to isolate x

$\Leftrightarrow x = \frac{2}{3}$ $<=>x=2/3$

and we have our answer

How do you solve 8x=48^x=4?