Inverse of f(x)=2^x-1

help:(

f(x) = 2^(x - 1) ?

let y = 2^(x - 1)
for inverse, x = 2^(y - 1)
log x = log [2^(y - 1)]
log x = (y - 1) log 2
log x = y log 2 - log 2
log x + log 2 = y log 2
(log x + log 2) / log 2 = y

y = (log x / log 2) + 1 <== inverse
or, y = (log_2 x) + 1 <== inverse

*****
f(x) = (2^x) - 1
y = (2^x) - 1
for inverse, x = (2^y) - 1
x + 1 = 2^y
log (x + 1) = log (2^y)
log (x + 1) = y log 2

y = log (x + 1) / log 2 or y = log_2 (x + 1) <== inverse

Set the equation as y = the desired function.
Then solve for x
Then when writing the solution replace the y with x.

y = 2^x - 1 add 1 to each side
y + 1 = 2^x take each side to the log_10 (log base 10)
log_10 (y + 1) = log _10 (2^x) = x log_10 (2) remember your rules of logs
log_10 (y + 1) = x log_10 (2) now divide by log_10 (2)
(log_10 (y + 1)) / (log _ 10(2)) = x

inverse of f(x) = (log_10 (x + 1)) / (log _10 (2))

f(x) is a fancy way of saying the other variable, let's say y.
y = 2^x - 1
Swap the variables
x = 2^y - 1
x+1 = 2^y
log(x+1) = y log(2)
(log(x+1)) / (log(2)) = y
Hell yeah, it's ugly but don't worry about it. You don't have to go further ;)

change f(x) to y

then swap x and y giving

x = (2^y)-1

solve that equation for y

x+1 = 2^y

log(base2)(x+1) = y = inverse of f(x)