# Finding the Inverse of an exponential Log

• Jul 11th 2012, 04:04 AM
dina
Finding the Inverse of an exponential Log
Hi I have a question on graphing this function and its inverse y =1/3 * e^(x+1) . In working this out I found that the inverse is X= ln(3y)-1 and hope this is correct
But I do not know how to draw both graph? Can someone assist me with this please?
• Jul 11th 2012, 04:33 AM
tom@ballooncalculus
Re: Finding the Inverse of an exponential Log
Find the location of crucial points e.g. where does the graph cross the axes, or does it? Maybe also, where is it highest and lowest, maybe even, which way is it curving at various points. Then join the dots accordingly, with a reasonably smooth curve. Check your sketch by using Graph - CNET Download.com

Also you want to check that your inverse is a reflection of the first. (In which line?)
• Jul 11th 2012, 07:34 AM
Soroban
Re: Finding the Inverse of an exponential Log
Hello, dina!

I assume that you are not expected to sketch the graphs
. . by calculating and plotting dozens (hundreds?) of points.

Quote:

Function: .$\displaystyle y \:=\:\tfrac{1}{3}e^{x+1}$
Inverse: .$\displaystyle y \:=\:\ln(3y)-1$

But I do not know how to draw both graphs.

You are expected to know the graph of the basic exponential function: .$\displaystyle y \:=\:e^x$

Code:

                |                 |      *                 |                 |    *                 |    *               1|  *                 *             *  |       *        |     - - - - - - + - - - - - - - -                 |
It has a y-intercept of (0,1).
To the left, it approaches the x-axis.
To the right, it rises rapidly.

We have: .$\displaystyle y \:=\:\tfrac{1}{3}e^{x+1}$

The $\displaystyle x\!+\!1$ moves the graph one unit to the left.

The $\displaystyle \tfrac{1}{3}$ compresses the graph vertically.
Each point is now a third as high.

Now we have the basic logarithmic function: .$\displaystyle y \:=\:\ln x$

Since it is the inverse of the exponential function,
. . its graph is the reflection over the 45o line.

Code:

        |        .         |      .         |    .                *         |  .          *         | .      *     - - + - -*- - - - - - - - - - -       . |  * 1     .  | *         |         |*         |
It has an x-intercept of (1,0).
To the left, it approaches the y-axis.
To the right, it rises slowly.

We have: .$\displaystyle y \:=\:\ln(3x) - 1$

The $\displaystyle 3x$ compresses the graph horizontally.
Each point is a third of its distance to the right.
. . The x-intercept is $\displaystyle \left(\tfrac{1}{3},\,0\right).$
Then the $\displaystyle \text{-}1$ lowers the entire graph one unit.