Question

Question #f34ef

Answer 1

`7^(2^x)=\approx2.08649`

Explanation:

`7^(-x)=4`
`1/7^x=4`
`1/4=7^x`
`log_(1/4)(7)=x`
`x\approx-1.40368`
I suggest you use the exact form (the logarithm) to calculate the next step, instead of the approximation.

`7^(2^x)=7^(2^(\log_(1/4)(7)))`

`\approx2.08649`