Inverse of f(x)= 2^3x + 1
To find the inverse of a function:
1. Replace f(x) with y.
2. Switch x and y.
3. Solve for y.
4. Replace y with f-1(x) (notation for inverse fucntions)
\(f(x)=2^{3x}+1\)
Let's use y instead of f(x).
\(y=2^{3x}+1\)
Switch the variables...
\(x=2^{3y}+1\)
Now, we solve for y.
\(x-1=2^{3y}\)
\(3y =log_2(x-1)\)
\(y=\frac{log_2(x-1)}{3}\)
Now, just put that inverse function notation on there.
\(\boxed{f^{-1}(x)=\frac{log_2(x-1)}{3}}\)
to find inverse
replace f(x) with y
solve for x
switch x and y and replace y with f(x)inverse or f⁻¹(x)
y=2^(3x)+1
subtract 1
y-1=2^(3x)
take log base 2 of both sides
log₂(y-1)=log₂(2^(3x))
move exponent (2x) to front
log₂(y-1)=3xlog₂(2)
remember that logₓ(x)=1 so
log₂(y-1)=3x(1)
log₂(y-1)=3x
divide both sides by 3
\(\frac{ log_{2}(y-1) }{3}\)=x
switch x and y and replace y with f⁻¹(x)
f⁻¹(x)=\(\frac{ log_{2}(x-1) }{3}\)