Stuck on how to solve simultaneous exponential equations

**angypangy** · January 16th 2012, 03:51 AM

Hello

I have the equations:

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

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

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

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

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

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

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

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

then

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

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

**princeps** · January 16th 2012, 04:10 AM

Originally Posted by angypangy

Hello

I have the equations:

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

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

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

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

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

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

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

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

then

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

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

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

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

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