Question

solve 3^(2x)=7^(x-1)

Answer 1

\[\Large\rm 3^{2x}=7^{x-1}\]Hey Annabelle :)
See how our x's are stuck up in the exponent?
Let's use logarithm to deal with that.
There a log rule that will help get them out of that place.

Answer 2

Let's take the natural log of each side,\[\large\rm \ln\left(3^{2x}\right)=\ln\left(7^{x-1}\right)\]

Answer 3

Here is the log rule that I have in mind:\[\large\rm \color{orangered}{\log(a^b)=b\cdot \log(a)}\]Do you understand how that will help us?

Answer 4

would the first step to multiply each side to balance the logs?

Answer 5

Hmmm, Im not sure what you mean :o

Answer 6

The first step was taking natural log of each side.
Next we apply our log rule to each side.

Answer 7

\[\left( 3^2 \right)^x=7^x*7^{-1},9^x=\frac{ 7^x }{ 7 },\frac{ 9^x }{ 7^x }=\frac{ 1 }{ 7 }\]
\[\left( \frac{ 9 }{ 7 } \right)^x=\frac{ 1 }{ 7 }\]
x[log9-log7]=log1-log7
x=?