Question

when you have a term like (-2^-x)ln2 and you want to bring the exponent down to make it positive how come its -(1/(2^x)(ln2) instead of -(ln2/(2^x))?

Answer 1

why does \[-2^{-x}\ln2\]=\[ -1\over 2^{x}\ln2\] and not
\[\ln2 \over 2^{x}\]?

Answer 2

because
\[2^{-x}=\frac{1}{2^x}\]

Answer 3

should be
\[-2^{-x}\ln(x)=-\frac{\ln(x)}{2^x}\]

Answer 4

I'm with you. ln2 is just a number.

So I think it should be \[-\ln 2/2^{x}\]

I'm curious where you saw it as \[-1/2^{x}\ln 2\]

Answer 5

sorry for the late response, but that was the answer in my textbook to a problem. I was thinking -ln2 / 2^x but the text says -1/2^x ln2 which is why I wanted to double check my algebra here

Answer 6

the problem was to find the integral of 2^-x

Answer 7

Since it's been so long since I took calculus.

Why don't post the intergal of 2^-x as a new question. Maybe there is something you are not seeing.

Answer 8

i just figured it out by looking at it again roflmao

Answer 9

it was the calculus, not the algebra, damn!