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?
$\endgroup$ 23 Answers
$\begingroup$$5*3^x = 2*7^x$ so
$\log (5*3^x) = \log(2*7^x)$
$\log 5 + \log 3^x = \log 2 + \log 7^x$
$\log 5 + x\log 3 = \log 2 + x \log 7$.
Now just treat those logs as constants.....
Hint:
$x \log 3 - x \log 7 = \log 2 - \log 5$
$x (\log 3 - \log 7) = \log 2 - \log 5$
$x = \frac {\log 2 - \log 5}{\log 3 - \log 7}$.
$\endgroup$ 3 $\begingroup$After taking $\log$ on both sides you actually get: $$\log 5+x\log3=\log 2+x\log 7$$ $$\Longrightarrow x= \frac{\log\frac{5}{2}}{\log \frac{7}{3}}$$
$\endgroup$ $\begingroup$A good start is always to collect the unknown: $$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)}.$$
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