Graphing help plz!

Question

For the exponential function f, find f^-1 analytically and graph both f and f^-1.

f(x)=5^x-1

Answer 1

When we say f(x) = 5^x - 1, what we're really saying is y = 5^x - 1. When we're trying to find f^-1, we're essentially trying to solve for x in terms of y and changing the y's back to x's. In this case, y = 5^x - 1 implies y+1 = 5^x . If we take log₅ of both sides, we get y in terms of x: log₅(y+1) = log₅(5^x) which implies x = log₅(y+1) or f^-1(x) = log₅(x+1).

Graphing is not something I can describe to you totally, but at x = 0 , y = 5^x - 1 is 0 and at 1, y = 5^x - 1 = 4. For log₅(x+1), if x = 0, log₅(x+1) = 0. If x = 4, log₅(x+1) = 1. The exponential line is essentially flat at y = -1 to the left of the x-axis, but the slope starts rising quicker and quicker after you get to the positive side of the x-axis. For the other function, the line is essentially vertical at x = -1 and slowly starts curving to the right as you get close to the y - axis until you get the y-axis, and then it starts curving more and more sharply to the right. That's the best I can do without giving a video answer.

Answer 2

f(x) = y = 5^x -1

y - 1 = 5^x

ln(y-1) = x* ln(5)

ln(y-1)/ln(5) = x

Now switch x and y....

f^-1(x) = ln(x-1)/ln(5)

With the functions f(x) and f^-1(x), you can plot them using desmos...