# Thread: Equations with the exponential function.

1. ## Equations with the exponential function.

Hello

I know this is elementary, but I don't know how solve this:

The question is, find the exact value(s) of x which satisfy the equation.

$e^{2x} = e^x +12$

I am not confident working with logarithms, although I do know the laws for the sum and difference of logarithms. I can do the other questions in this exercise, but not this one.

I know if the question was:

$e^{2x} = 12$

$x = \dfrac{1}{2}\ln12$

Some help would be very much appreciated. Thank you.

2. ## Re: Equations with the exponential function.

Originally Posted by Furyan
Hello

I know this is elementary, but I don't know how solve this:

The question is, find the exact value(s) of x which satisfy the equation.

$e^{2x} = e^x +12$

I am not confident working with logarithms, although I do know the laws for the sum and difference of logarithms. I can do the other questions in this exercise, but not this one.

I know if the question was:

$e^{2x} = 12$

$x = \dfrac{1}{2}\ln12$

Some help would be very much appreciated. Thank you.
$e^{2x} - e^x - 12 = 0$

$(e^x - 4)(e^x + 3) = 0$

finish it

3. ## Re: Equations with the exponential function.

Dear Skeeter

Originally Posted by skeeter
$e^{2x} - e^x - 12 = 0$

$(e^x - 4)(e^x + 3) = 0$

finish it
A quadratic in $e$, that was left field and I wasn't looking for it. Should have seen it though.

$x = \ln4 = 2\ln2$

$e^x = -3$ has no solutions.

Thank you very much indeed.

4. ## Re: Equations with the exponential function.

Hopefully, you know that $e^{2x}= (e^x)^2$. If you let $y= e^x$, the equation becomes $y^2= y+ 12$ which is the same as $y^2- y- 12= 0$, a quadratic equation in y. If you did not notice that this could be factored as $(y- 4)(y+ 3)$, you could solve it by completing the square or using the quadratic formula.

5. ## Re: Equations with the exponential function.

Hello HallsofIvy

Originally Posted by HallsofIvy
Hopefully, you know that $e^{2x}= (e^x)^2$. If you let $y= e^x$, the equation becomes $y^2= y+ 12$ which is the same as $y^2- y- 12= 0$, a quadratic equation in y. If you did not notice that this could be factored as $(y- 4)(y+ 3)$, you could solve it by completing the square or using the quadratic formula.
Thank you. I do find letting $y= e^x$ and solving the quadratic in $y$ easier to get my head round. I was able to factor that one.

6. ## Re: Equations with the exponential function.

Hello, Furyan!

Find the exact value(s) of $x$ which satisfy the equation."

. . $e^{2x} - e^x -12\:=\:0$

As skeeter pointed out, this is a quadratic equation.

There is a way to recognize this phenomenon.

There are usually three terms, written in standard order.

If the first term has twice the exponent of the second term,
. . we may have a quadratic.

In your problem: . $e^{2(x)} - e^x - 12 \:=\:0$

Another example: . $x^6 - 7x^3 - 8 \:=\:0$

Note that we have: . $\left(x^3\right)^2 - 7(x^3) - 8 \:=\:0$

Let $u \,=\,x^3$

Then we have: . $u^2 - 7u - 8 \:=\:0$
. . $(u-8)(u+1) \:=\:0 \quad\Rightarrow\quad u \:=\:8,\text{-}1$

Back-substitute: . $\begin{Bmatrix}x^3 \:=\:8 & \Rightarrow & x \:=\:2 \\ x^3 \:=\:\text{-}1 & \Rightarrow & x \:=\:\text{-}1 \end{Bmatrix}$

And there are four complex roots as well.

7. ## Re: Equations with the exponential function.

Hello Soroban

Originally Posted by Soroban
Hello, Furyan!

As skeeter pointed out, this is a quadratic equation.

There is a way to recognize this phenomenon.

There are usually three terms, written in standard order.

If the first term has twice the exponent of the second term,
. . we may have a quadratic.

In your problem: . $e^{2(x)} - e^x - 12 \:=\:0$

Another example: . $x^6 - 7x^3 - 8 \:=\:0$

Note that we have: . $\left(x^3\right)^2 - 7(x^3) - 8 \:=\:0$

Let $u \,=\,x^3$

Then we have: . $u^2 - 7u - 8 \:=\:0$
. . $(u-8)(u+1) \:=\:0 \quad\Rightarrow\quad u \:=\:8,\text{-}1$

Back-substitute: . $\begin{Bmatrix}x^3 \:=\:8 & \Rightarrow & x \:=\:2 \\ x^3 \:=\:\text{-}1 & \Rightarrow & x \:=\:\text{-}1 \end{Bmatrix}$

And there are four complex roots as well.

Thank you for that. I will be sure to look out for this sort of equation in the future.