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This one is an exponential equation that I can't figure out..

$7^{x-2} = 5^{3-x}$

These two are logarithmic equations that I'm also having trouble with..

$\ln \sqrt[3]{x-6} = -2$

$\displaystyle\frac{1025}{7+e^{4x}} = 5$

These ones really stumped me. Any explanations would be greatly appreciated! Thank you!

wckronholm
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Jordan
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  • $(x-2)\log_{10} 7 = (3-x)\log_{10} 5$, etc. – Michael Hardy Feb 02 '14 at 18:23
  • How do you proceed from there, though? That's where I'm stuck with that one. – Jordan Feb 02 '14 at 18:27
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    It's a linear equation. More specifically, $\displaystyle x \log_{10} 7 - 2 \log_{10} 7 = 3 \log_{10} 5 - x \log_{10} 5$. Gather the $x$ terms together to one side and the constants to the other, do some algebra, and voila! – 2012ssohn Feb 02 '14 at 18:29

2 Answers2

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Hint for #1: use Michael Hardy's suggestion.

Hint for #2: exponentiate both sides, using the fact that $\displaystyle e^{\ln x} = x$.

Hint for #3: bring the $e^{4x}$ term to the numerator, isolate it, then take the natural log, using the fact that $\displaystyle \ln e^x = x$.

2012ssohn
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$$\log\sqrt[3]{x-6}=-2\iff \log(x-6)=-6\iff x-6=e^{-6}\;\ldots$$

$$\frac{1,025}{7+e^{4x}}=5\iff5e^{4x}=990\iff e^{4x}=198\iff 4x=\log198\;\ldots$$

DonAntonio
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  • I would guess that someone who didn't know what to do with the equation involving the radical failed to realize that $\log\sqrt[3]{x-6}$ is $\frac13\log(x-6)$, so I'd have said that explicitly. – Michael Hardy Feb 02 '14 at 18:35
  • @MichaelHardy I just suggested that as a reminder since imo someone who has to deal with these problems must know the basic properties of logarithms... – DonAntonio Feb 02 '14 at 18:39
  • Since they MUST know them, you should not explicitly inform them when they are found not to know them? – Michael Hardy Feb 02 '14 at 18:46
  • Ok, that was too much for me, @MichaelHardy: I just couldn't understand what you mean. – DonAntonio Feb 02 '14 at 18:48
  • I do know the properties of logarithms, and I just solved the equation. I roughly got x = 6.002 .. The answer for that problem was extremely helpful. I just needed a hint on how to start, & then it all made sense. Thank you! – Jordan Feb 02 '14 at 18:53
  • @DonAntonio : I said I'd have mentioned it explicitly in the answer rather than writing what you wrote. You responded with something I couldn't understand plus the statement that someone who has to do problems like this MUST know the properties of logarithms. Did you mean that the fact that they MUST know them is a reason why the answer should not state them explicitly? – Michael Hardy Feb 02 '14 at 18:58
  • Also, unfortunately most textbooks at this level introduce an explicit convention that $\log x$ means $\log_{10} x$, not $\log_e x$. – Michael Hardy Feb 02 '14 at 18:59