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

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Answer 1 · 2017-01-17T00:58:49

#8^x=4 <=> x=2/3#

#8^x=4#

Since #4=(2)(2)=2^2#

we can say that

#8=4(2)=(2)(2)(2)=2(2^2)=2^(2+1)=2^3#

So by changing notation we get

#<=> (2^3)^x=2^2#

and since #(x^a)^b=x^(ab)#, we can say

# <=> 2^(3x)=2^2#

Then we take the #ln_2(x)# of both sides

#<=> ln_2(2^(3x))=ln_2(2^2)#

This gives us

# <=> 3x=2#

Then we divide both sides by 3 to isolate x

# <=>x=2/3#

and we have our answer