# Solve the equation

• Feb 25th 2013, 11:30 AM
Paze
Solve the equation
Hi MHF.

I am asked to solve the equation $3^{2x-1}+27=10\cdot3^x$

I can pretend that $3^x=t$ and solve for $t$ to get $t1=27\\t2=3$ however the book states that $x1=1\\x2=3$ I don't know how to solve for x.. Only t..So far !

Perhaps MHF can clear this one up for me! Also, am I in the right section with this problem, or is this simply algebra? Thank you!
• Feb 25th 2013, 11:55 AM
HallsofIvy
Re: Solve the equation
You were the one who decided to make $t= 3^x$! And once you found that t= 27 and t= 1 are roots, you should immediately realize that $3^x= 27$ and $3^x= 3$. Isn't it clear what x must be?

(If it were something more complicated, say $3^x= 25$, you could use a logarithm: $log(3^x)= xlog(3)= log(25)$ and then $x= \frac{log(25)}{log(3)}$ where the logarithm can be to any convenient base. But since both "3" and "27" are powers of 3, you were probably not intended to do that.)
• Feb 25th 2013, 11:58 AM
Paze
Re: Solve the equation
Quote:

Originally Posted by HallsofIvy
You were the one who decided to make $t= 3^x$! And once you found that t= 27 and t= 1 are roots, you should immediately realize that $3^x= 27$ and $3^x= 3$. Isn't it clear what x must be?

(If it were something more complicated, say $3^x= 25$, you could use a logarithm: $log(3^x)= xlog(3)= log(25)$ and then $x= \frac{log(25)}{log(3)}$ where the logarithm can be to any convenient base. But since both "3" and "27" are powers of 3, you were probably not intended to do that.)

Oh my. I didn't realize that. The answer was right in front of me! I guess I am slowly moving into the logarithm method you posted so thanks for that too!
• Feb 25th 2013, 02:21 PM
Soroban
Re: Solve the equation
Hello, Paze!

Quote:

$\text{Solve: }\:3^{2x-1}+27\:=\:10\cdot3^x$

Multiply by 3: . $3^{2x} + 81 \:=\:30\cdot3^x$

Let $t = 3^x$
Then we have: . $t^2 + 81 \:=\:30t \quad\Rightarrow\quad t^2 - 30t + 81 \:=\:0$

. . $(t-3)(t-27) \:=\:0 \quad\Rightarrow\quad \begin{Bmatrix}t \,=\,3\\t\,=\,27\end{Bmatrix}$

Back-substitute: . $\begin{Bmatrix}3^x \:=\:3 & \Rightarrow& x \:=\:1 \\ 3^x \:=\:27 & \Rightarrow & x \:=\:3 \end{Bmatrix}$