# Solve the Logarithmic Equations for x, please.

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!

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$(x-2)\log_{10} 7 = (3-x)\log_{10} 5$, etc. –  Michael Hardy Feb 2 '14 at 18:23
How do you proceed from there, though? That's where I'm stuck with that one. –  Jordan Feb 2 '14 at 18:27
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 2 '14 at 18:29

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 $\displaystyle \ln e^x = x$.

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

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

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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 2 '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 2 '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 2 '14 at 18:46
Ok, that was too much for me, @MichaelHardy: I just couldn't understand what you mean. –  DonAntonio Feb 2 '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 2 '14 at 18:53