# Taking e to both sides [Solved!]

**phinah** 22 Apr 2018, 16:18

### My question

Your Integration Chapter: The Basic Logarithmic Form, example 4.

Can you show how you took e to both sides to arrive at your result please?

### Relevant page

2. Integration: The Basic Logarithm Form

### What I've done so far

After integrating, `t = ln 20 - ln (20-v)`.

You applied log laws to get `t = ln(20/(20-v))`.

You took "e to both sides" to get `e^t = 20/20-v`.

We thought by taking e to both sides the result would be the exponents equated:

`t = 20/20-v`.

X

Your Integration Chapter: The Basic Logarithmic Form, example 4.
Can you show how you took e to both sides to arrive at your result please?

Relevant page
<a href="https://www.intmath.com/methods-integration/2-integration-logarithmic-form.php">2. Integration: The Basic Logarithm Form</a>
What I've done so far
After integrating, `t = ln 20 - ln (20-v)`.
You applied log laws to get `t = ln(20/(20-v))`.
You took "e to both sides" to get `e^t = 20/20-v`.
We thought by taking e to both sides the result would be the exponents equated:
`t = 20/20-v`.

## Re: Taking e to both sides

**Murray** 23 Apr 2018, 02:48

@Phinah: The opposite process of taking the "log" of something is to take "e to the power of...".

Writing the full details would be:

`t = ln (20 /(20-v))`

`e^t = e^[ln (20/(20-v))]`

`e^t = 20/(20-v)`

If we had

`e^t = e^[(20/(20-v))]` (without "`ln`")

then the answer would be

`t = 20/(20-v)`

but that's not what we started with.

A similar situation is the case where "raising to power 2" is the opposite of "square root".

So let's start with

`y = sqrt(x+2)`

Squaring both sides gives us:

`y^2 = (sqrt(x+2))^2 = x+2`

The "power 2" has "undone" the `sqrt` operation.

Similarly, the "`e` to the power" operation undoes the "`ln`" expression in the example you are asking about.

Hope it makes sense.

X

@Phinah: The opposite process of taking the "log" of something is to take "e to the power of...".
Writing the full details would be:
`t = ln (20 /(20-v))`
`e^t = e^[ln (20/(20-v))]`
`e^t = 20/(20-v)`
If we had
`e^t = e^[(20/(20-v))]` (without "`ln`")
then the answer would be
`t = 20/(20-v)`
but that's not what we started with.
A similar situation is the case where "raising to power 2" is the opposite of "square root".
So let's start with
`y = sqrt(x+2)`
Squaring both sides gives us:
`y^2 = (sqrt(x+2))^2 = x+2`
The "power 2" has "undone" the `sqrt` operation.
Similarly, the "`e` to the power" operation undoes the "`ln`" expression in the example you are asking about.
Hope it makes sense.

## Re: Taking e to both sides

**phinah** 25 Apr 2018, 10:07

It does now. Thank you.

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