Logarithmic functions

Question

Given f(x) = 4(2)^x . a. Find f-1(x). Note: Your function should be a single logarithm so use the change of base formula if needed.

b. First graph f(x) dotted and then use inverses to graph f-1(x) solid below.

c. State the equation of the asymptote for each graph. Asymptote for f(x) ______________

Asymptote for f-1(x) ______________

Answer 1

To find the inverse of a function, write the function as "y = " instead of "f(x) = " then just switch the x and y around and solve for y. The resulting "y =" function will be the inverse of f(x). Like this:

f(x) = 4(2)^x

y = 4(2)^x

x = 4(2)^y

(1/4)x = (2)^y

log₂(1/4x) = log₂(2)^y

log₂(1/4) + log₂(x) = ylog₂(2)

log₂(2^-2) + log₂(x) = y(1)

-2log₂(2) + log₂(x) = y

y = log₂(x) - 2

f^-1(x) = log₂(x) - 2

The graphs would look like this:

For f(x) there is a horizontal asymptote at y = 0 (the x-axis)

For f^-1(x) there is a vertical asymptote at x = 0 (the y-axis)

Answer 2

y = 4(2^x), where y = f(x)

Switch x and y: x = 4(2^y)

Solve for y: (x/4) = 2^y. So, log₂(x/4) = y

y = log₂x - log₂4

y = f^-1(x) = log₂x - 2

y = f(x) has the x-axis as horizontal asymptote and has no vertical asymptote

y = f^-1(x) has the y-axis as vertical asymptote, but doesn't have a horizontal asymptote

The graph of y = f^-1(x) can be obtained from the graph of y = f(x) by reflecting the graph of y = f(x) through the line y = x.