Question

How do I solve 5^(-2x)=3?

Answer 1

change of base says
\[b^x-A\iff x=\frac{\ln(A)}{\ln (b)}\]

Answer 2

make that
\[b^x=A\iff x=\frac{\ln(A)}{\ln (b)}\]

Answer 3

Thanks!

Answer 4

you probably do not have log base 5 on your calculator, which is why you need the change of base formula. in one step you should write
\[-2x=\frac{\ln(3)}{\ln(5)}\] and so \[x=\frac{\ln(3)}{-2\ln(5)}\]

Answer 5

Awesome! Thanks--That's just what I needed :)

Answer 6

yw