# Stuck on how to solve simultaneous exponential equations

• Jan 16th 2012, 03:51 AM
angypangy
Stuck on how to solve simultaneous exponential equations
Hello

I have the equations:

$\displaystyle 6 = a2^n$ and $\displaystyle 18 = a3^n$

I am really stuck on this. This is what I tried:

$\displaystyle 2^n = \frac{6}{a}$

$\displaystyle 3^n = \frac{18}{a}$

$\displaystyle n log2 = log \frac{6}{a} = log(6) - log(a)$

$\displaystyle n log3 = log \frac{18}{a} = log(18) - log(a)$

$\displaystyle n = \frac{log(6) - log(a)}{log2}$

$\displaystyle n = \frac{log(18) - log(a)}{log3}$

then

$\displaystyle \frac{log(6) - log(a)}{log2} = \frac{log(18) - log(a)}{log3}$

Is this the way to solve this sort of problem? Or am I missing an easier way?

If this is the way, it looks complicated. How would I proceed from here?

Angus
• Jan 16th 2012, 04:10 AM
princeps
Re: Stuck on how to solve simultaneous exponential equations
Quote:

Originally Posted by angypangy
Hello

I have the equations:

$\displaystyle 6 = a2^n$ and $\displaystyle 18 = a3^n$

I am really stuck on this. This is what I tried:

$\displaystyle 2^n = \frac{6}{a}$

$\displaystyle 3^n = \frac{18}{a}$

$\displaystyle n log2 = log \frac{6}{a} = log(6) - log(a)$

$\displaystyle n log3 = log \frac{18}{a} = log(18) - log(a)$

$\displaystyle n = \frac{log(6) - log(a)}{log2}$

$\displaystyle n = \frac{log(18) - log(a)}{log3}$

then

$\displaystyle \frac{log(6) - log(a)}{log2} = \frac{log(18) - log(a)}{log3}$

Is this the way to solve this sort of problem? Or am I missing an easier way?

If this is the way, it looks complicated. How would I proceed from here?

Angus

$\displaystyle a \cdot 3^n=6 \cdot 3 \Rightarrow a \cdot 3^n=a \cdot 2^n \cdot 3 \Rightarrow 3^n = 2^n \cdot 3 \Rightarrow \left(\frac{3}{2}\right)^n=3$

$\displaystyle \Rightarrow n=\log_{\frac{3}{2}} 3 ; a \neq 0$