how do I solve for x when it is an exponent on both sides?

Question

I am given the equation:

5 * 3^x = 2 * 7^x

The text book and everywhere online shows me how to do this when the variable is only on one side or when it can be subtracted/added, not when it is tied to a multiplication.

Of course the latter is going to be on the test and not what is taught in the book.

How am I supposed to edit this to get rid of the two down votes? Was this question already asked? What could've I possibly typed in to get my answer?

Answer 1

$5*3^x = 2*7^x$ so

$\log (5*3^x) = \log(2*7^x)$

$\log 5 + \log 3^x = \log 2 + \log 7^x$

$\log 5 + x\log 3 = \log 2 + x \log 7$.

Now just treat those logs as constants.....

Hint:

$x \log 3 - x \log 7 = \log 2 - \log 5$

$x (\log 3 - \log 7) = \log 2 - \log 5$

$x = \frac {\log 2 - \log 5}{\log 3 - \log 7}$.

Answer 2

After taking $\log$ on both sides you actually get: $$\log 5+x\log3=\log 2+x\log 7$$ $$\Longrightarrow x= \frac{\log\frac{5}{2}}{\log \frac{7}{3}}$$

Answer 3

A good start is always to collect the unknown: $$5=2\cdot\frac{7^x}{3^x}\iff5=2\cdot\left(\frac73\right)^x$$ an then isolate: $$5/2=(7/3)^x.$$ Now take logarithm $$\ln(5/2)=x\cdot\ln(7/3)$$ and arrive in $$x=\frac{\ln(5/2)}{\ln(7/3)}.$$