# Thread: Need some Help with an Exponential Equation

1. ## Need some Help with an Exponential Equation

Thanks in advance for any help!

I am having trouble with this equation:

2^(3*x)-3*2^(2*x)-6*2^x+8=0

My main problem is how to deal with the exponents that are multiplied by a constant, here:

2^(3*x)-3*2^(2*x)-6*2^x+8=0

I cant just multiply these can I ? and end up with 12^x and so on, can I?

Thank you!

2. ## Re: Need some Help with an Exponential Equation

Let $2^{x}=X$ and you will find that you have a cubic equation in $X.$
One solution of the equation is easy to spot and you can then factorise.

3. ## Re: Need some Help with an Exponential Equation

I'm sorry but I'm not sure I understand your answer? I should replace the power, 2x with X? So can I just multiply the constants in front of the powers by the constants, then?

Thanks again!

4. ## Re: Need some Help with an Exponential Equation

The second term, for example, $3*2^{2x}$ can be rewritten as $3*(2^{x})^{2}=3X^{2}.$
Substitute in the first and third terms in the same way and you have a cubic equation in $X.$

5. ## Re: Need some Help with an Exponential Equation

Hello, Rhys101!

$2^{3x} - 3\cdot2^{2x} - 6\cdot2^x+8\:=\:0$

We have: . $(2^x)^3 - 3(2^x)^2 - 6(2^x) + 8 \:=\:0$

Let $u = 2^x\!:\;\;u^3 - 3u^2 - 6u + 8 \:=\:0$

Factor: . $(u-1)(u+2)(u-4) \:=\:0$

We have three roots: . $u \;=\;1,\,\text{-}2,\,4$

Back-substitute: . $\begin{Bmatrix}2^x \,=\,1 & \Rightarrow & x \,=\,0 \\ 2^x \,=\,\text{-}2 & \Rightarrow & \text{no real root} \\ 2^x \,=\,4 & \Rightarrow & x \,=\,2 \end{array}$

Therefore: . $x\;=\;0,\,2$

6. ## Re: Need some Help with an Exponential Equation

Thanks so much for all your help, both of you! I really appreciated it!