Is this an unsolvable LOG equation?

Nov 2016
42
1
USA
Solve: log (x-10) = 3 + log (x-3)

Seems that no matter what is substituted for x, the log (x-10) is never able to equal the log (x-3) plus 3. Is this equation unsolvable, or are there theoretical/imaginary solutions?
 

romsek

Math Team
Sep 2015
2,762
1,541
USA
I'm assuming $\log(x) \sim \log_{10}(x)$

$\log(x-10)=3+\log(x-3)\\

\log(x-10)-\log(x-3) = 3 \\

\log\left(\dfrac{x-10}{x-3}\right) = 3\\

\left(\dfrac{x-10}{x-3}\right) = 10^3 = 1000\\

x-10 = 1000(x-3) = 1000x - 3000\\

999x = 2990\\

x = \dfrac{2990}{999}$
 
Dec 2015
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Earth
For any equation you must be able to spot \(\displaystyle 3=\log10^3\).
Then remove the logs both sides .
 
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greg1313

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Oct 2008
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Assuming base 10 logs,

\(\displaystyle \log(x-10)=3+\log(x-3) \)

\(\displaystyle \log(x-10)-\log(x-3)=3\)

\(\displaystyle \log\left(\frac{x-10}{x-3}\right)=3\quad(*)\)

\(\displaystyle \frac{x-10}{x-3}=1000\)

\(\displaystyle x-10=1000x-3000\)

\(\displaystyle 999x=2990\)

\(\displaystyle x=\frac{2990}{999}\)

If you use $(*)$ to verify, you avoid complex numbers.

Posted $\sim$ romsek.
 
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Nov 2016
42
1
USA
Is x=2990/999 (2.993) an imaginary answer? Plug in 2.993 into the left side of original equation (log x-10):

log (2.993-10)

= log (-7.007), but this seems unsolvable?
 
Jun 2019
493
260
USA
2990/999 is a real number, but the logs are both complex numbers. So, you can only find the answer using algebra, or using a numeric environment that handles complex numbers.
 
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romsek

Math Team
Sep 2015
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I should have gone back and verified that the solution didn't break any rules, which it does.
But OP needs to specify whether we are talking strictly reals or not.

Given it's high school math it almost certainly is.
 
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Dec 2015
935
122
Earth
\(\displaystyle re^{i\pi}=-r \; \) ; \(\displaystyle \; \ln(-r)=\ln(r)+i\pi\).

\(\displaystyle \log(-7.007 )=\dfrac{\ln(-7.007 )}{\ln10} =\dfrac{\ln(7.007)}{\ln10}+i\dfrac{\pi}{\ln10}\).
 
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skipjack

Forum Staff
Dec 2006
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If "log" in the original post is the natural logarithm, there is a real solution.
 
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