Question

Suppose a and b are real numbers such that 17^a=16 and 17^b=4. What is 1/(2^a-b) ?

Answer 1

Since

`17^a = 16 = 4^2 \implies 17^(a/2) = 4`

it follows that

`17^b = 4 \implies 17^(a/2) = 17^b \implies \frac a2 = b \implies a = 2b`

Then

`2^(a - b) = 2^(2b - b) = 2^b`

so that

`\frac1{2^(a-b)} = 2^(-b)`

We also have

`17^b = 4 \implies b = \log_(17)(4)`

so we can go on to say

`\frac1{2^(a-b)} = \boxed{2^{-\log_(17)(4)}}`