rewrittenOur original equation:
[tex](\frac{1}{2})^x=\text{ 8}[/tex]
We need to find the value of x, so if x>1 the result will never get 8 because the exponent:
[tex](\frac{1}{2})^6=\frac{(1)^6}{(2)^6}=\frac{1}{64}[/tex]
So, let's use a negative exponent:
In this case -3, because 2^3= 2*2*2=8.
[tex](\frac{1}{2})^{-3}=\frac{(1)^{-3}}{(2)^{-3}}=\frac{\frac{1}{1^3}}{\frac{1}{2^3}}[/tex]
Operate and solve:
[tex]\frac{\frac{1}{1}}{\frac{1}{8}}=\frac{8}{1}=8[/tex]
So, rewriten equation:
[tex](\frac{1}{2})^x=(\frac{1}{2})^{-3}[/tex]
Solution:
x= -3
