Respuesta :
Answer: s=-1
Explanation:
We can re-write the equation as follows:
[tex]\frac{1}{64}=(4^{2s-1})(16^{2s+2})[/tex]
Let's rewrite the terms on the left and on the right as follows:
[tex]2^{-6} = ((2^2)^{2s-1})((2^4)^{2s+2})[/tex]
So we have
[tex]2^{-6} = (2^{2(2s-1)})(2^{4(2s+2)})[/tex]
Now we can use the following rule:
[tex]x^{a+b}=x^a x^b[/tex]
to rewrite the term on the right:
[tex](2^{2(2s-1)})(2^{4(2s+2)})=2^{2(2s-1)+4(2s+2)}[/tex]
So now we have
[tex]2^{-6}=2^{2(2s-1)+4(2s+2)}[/tex]
and since the bases are the same, we can just equalize the exponents:
[tex]-6=2(2s-1)+4(2s+2)[/tex]
[tex]-6=4s-2+8s+8\\-12 = 12 s\\s=-1[/tex]