Question

How do you write the inverse of h(x) = e 2x

h -1(x) =
a)loge2x
b)2logex
c)½logex

Answer 1

ideally, you sqap out the x and h(x) parts and resolve in for h(x)

Answer 2

not too sure what "e 2x" means tho

Answer 3

*swap

Answer 4

\[\huge e^{2x} = y\]now solve for x
log base e both sides\[\huge 2x = \log_e y\]and then yea
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Answer 5

Well thank y'all that helps a lot