# Solve for x: `41 - e^(2x) = 5`

ishpiro | College Teacher | (Level 1) Educator

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`41 - e^(2x) = 5`

In order to solve this equation for x, first isolate the exponential expression. To do that, subtract 41 from both sides"

`-e^(2x) = -36`

Divide by (-1) to eliminate the negative sign on both sides:

`e^(2x) = 36`

To solve for x, take the natural logarithm of both sides. Remember that natural logarithm of

`e^(2x)`

will be 2x.

`ln(e^(2x)) = ln(36)`

2x = ln(36)

Dividing both sides of the equation by 2, we get

`x = (ln(36))/2`

This can be further simplified by using the properties of logarithms. Since 36 can be written as `36 = 6^2` , then

`ln36 = ln6^2 = 2ln6`

So, `x = (2ln6)/2 = ln6`

The answer is `x= ln6` , which is approximately 1.79.

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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The value of x has to be found that satisfies the equation `41 - e^(2x) = 5` .

`41 - e^(2x) = 5`

`41 - 5 - e^(2x) + e^(2x) = 5 - 5 + e^(2x)`

`36 = e^(2x)`

Now 36 has to be written in the form `e^a` , use the relation` a^(log_a b) = b`

`36 = e^(ln 36)`

`e^(ln 36) = e^(2x)`

`ln 36 = 2x`

Now use the relation `log a^b = b*log a`

`2*ln 6 = 2x`

`x = ln 6`

The solution of the equation is x = ln 6

lochana2500 | Student, Undergraduate | (Level 1) Valedictorian

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`41` - `e^(2x)` `=` `5`

`e^(2x)` ` ` `=` `36`

`lne^(2x)` `=ln36`

` 2x = ln 6^(2)`

` `

` ` `2x= 2 ln6 `

`x= ln6 `

rachellopez | Student, Grade 12 | (Level 1) Valedictorian

Posted on

Your first goal is to isolate the e.

41-e^(2x)=5

-e^(2x)=-36

e^(2x)=36

In order to eliminate the e, you have to take the opposite of the action, which is a natural log, or ln.

ln(e^(2x))=ln36

2x=ln36

Now you can just solve for x.

x=ln36/2

You can simplify this by writing 36 as 6^2 and using the property of logarithms that allows you to move the exponent to the front.

x=ln6^(2)/2

x=2ln6/2

The 2s cancel and you are left with x=ln6