1. ## Solve exponential equation

Hi,

I'm currently trying to solve for the following exponential equation: 3^(2x)-5(3^x)=-6

I'm hoping that someone can explain why the steps I took to solve this equation are incorrect:
line 1 - 3^(2x)-5(3^x)=-6
line 2 - 3^(2x)-15^x=-6
line 3 - 3^x(1^x-5) = -6
line 4 - 3^x(-4)=-6
line 5 - 3^x=3/2
line 6 - xlog3=log(3/2)
line 7 - x=(log(3/2)/(log3)
line 8 - x=0.37...

The reason why I thought it was okay to remove the variable exponent from 1 after line 3 is because any exponent to 1 is just one. Which lead me to just subtract 5 from 1.

I am also familiar with another method of solving this, which is by letting 3^x equal a variable (e.g. let 3^x = u) and then solve the problem as if it is a quadratic equation. This method lead me to the correct answer. However, I still don't understand why the previous method does not work.

2. ## Re: Solve exponential equation

I'm currently trying to solve for the following exponential equation: 3^(2x)-5(3^x)=-6
$3^{2x} - 5 \cdot 3^x + 6 = 0$

let $t = 3^x$ ...

$t^2 - 5t + 6 = 0$

solve for $t$, then set each solution equal to $3^x$ and solve for $x$ ... remember $3^x > 0$ for all $x$

3. ## Re: Solve exponential equation

In line 2, you've written $5 \times (3^x)$ as $15^x$ which is not true. A simple counter example would be for $x=2$, $5 \times 3^2 = 5 \times 9=45$ whereas $15^2 = 225$.

Also, $3^{2x} \neq 3^x \times 1^x$, it is $3^{2x} = 3^x \times 3^x=\left(3^x\right)^2$

4. ## Re: Solve exponential equation

Originally Posted by Ahri
In line 2, you've written $5 \times (3^x)$ as $15^x$ which is not true. A simple counter example would be for $x=2$, $5 \times 3^2 = 5 \times 9=45$ whereas $15^2 = 225$.

Also, $3^{2x} \neq 3^x \times 1^x$, it is $3^{2x} = 3^x \times 3^x=\left(3^x\right)^2$
Oh right you have a good point there. Plus, I now just remembered that the laws of exponents can only be applied to numbers with the same base. Thanks

5. ## Re: Solve exponential equation

Originally Posted by otownsend
Oh right you have a good point there. Plus, I now just remembered that the laws of exponents can only be applied to numbers with the same base. Thanks
Look at reply #2. You must solve $t^2-5t+6=0$ where $t=3^x$.
(t-2)(t-3)=0 or $\left\{ \begin{array}{l}t=2~ \Rightarrow~3^x=2~ \Rightarrow~x=\dfrac{\log(2)}{\log(3)}\\t=3~ \Rightarrow~3^x=3~ \Rightarrow~x=1 \end{array} \right.$