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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?

Jason_Silva
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  • I know how to take the log of both sides You may want to give that a second try, since $\log(5 \cdot 3^x) \ne 5 \cdot x \cdot \log(3)$. – dxiv Apr 02 '18 at 05:35
  • "I know how to take the log of both sides and end up with:

    5 * x * log(3) = 2 * x * log(7)". That ISN'T what you should end up with when you take the logs from both sides. Taking the logs from both sides IS what you should do. But you should do it correctly; not incorrectly.

     – fleablood Apr 02 '18 at 05:41

3 Answers3

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

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

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

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

fleablood
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  • If I ended up with log 5 - log 2 = x log 7 - x log 3 after you said "Hint: ..." would it have been wrong? If so, then how am I supposed to know what to move over to which side? – Jason_Silva Apr 02 '18 at 16:51
  • No that would have been fine. You'd be saying $-A = -B$ instead of $A=B$. Try to think about what equations mean rather than about rules to "move things over". – Ethan Bolker Apr 02 '18 at 17:18
  • That would not have been in the least bit wrong. It doesn't matter what side you move things over as long as you do it correctly. If you had $A + Bx = C + Dx$ and I did $Bx - Dx = C - A$ and you did $A-C = Dx -Bx$ and I continued and got $x(B-D) = C-A$ and you cont. and got $A-C = x(D-B)$ and I concluded $x=\frac{C-A}{B-D}$ and you concluded $\frac {A-C}{D-B} = x$. Which of us is right and which of us is wrong? Well, unless $D=B$ then $\frac {A-C}{D-B} = \frac {-(C-A)}{-(B-D)} = \frac {C-A}{B-D}$ and we are both right and neither wrong. (Of course, if $B=D$ then we were both wrong....) – fleablood Apr 02 '18 at 19:04
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After taking $\log$ on both sides you actually get: $$\log 5+x\log3=\log 2+x\log 7$$ $$\Longrightarrow x= \frac{\log\frac{5}{2}}{\log \frac{7}{3}}$$

Jim Haddocc
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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 $$x=\frac{\ln(5/2)}{\ln(7/3)}.$$

Michael Hoppe
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