Question

If 1/64=4^2s-1 times 16^2s+2 what is the value of s?

Answer 1

s=-1 see attachment for explanation

Answer 2

Answer: s=-1

Step-by-step explanation:

We can re-write the equation as follows:

`(1)/(64)=(4^(2s-1))(16^(2s+2))`

Let's rewrite the terms on the left and on the right as follows:

`2^(-6) = ((2^2)^(2s-1))((2^4)^(2s+2))`

So we have

`2^(-6) = (2^(2(2s-1)))(2^(4(2s+2)))`

Now we can use the following rule:

`x^(a+b)=x^a x^b`

to rewrite the term on the right:

`(2^(2(2s-1)))(2^(4(2s+2)))=2^(2(2s-1)+4(2s+2))`

So now we have

`2^(-6)=2^(2(2s-1)+4(2s+2))`

and since the bases are the same, we can just equalize the exponents:

`-6=2(2s-1)+4(2s+2)`

`-6=4s-2+8s+8\\-12 = 12 s\\s=-1`