12^x-2=20

Is the 12 being raised to the power of "x" or the power of (x-2) ??

my answer is 24

I'll answerthe question asked but you do need to use brackets AllSmilesAlways to make your meaning clear.

$$\begin{array}{rll} 12^x-2&=&20\\\\ 12^x&=&22\\\\ log(12^x)&=&log22\\\\ xlog12&=&log22\\\\ x&=\frac{log22}{log12}\\\\ x&\approx &1.244 \mbox{ correct to 3 decimal places} \end{array}$$

Can someone please tell me why this did not align properly?

\begin{array}{rll}

12^x-2&=&20\\\\

12^x&=&22\\\\

log(12^x)&=&log22\\\\

xlog12&=&log22\\\\

x&=&\frac{log22}{log12}\\\\

x&\approx &1.244 \mbox{ correct to 3 decimal places}

\end{array}

You missed the & in front of \frac.

Thanks you Bertie.