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So, I have this logarithmic expression $$\log_5 8-\log_5 20-\log_5 10$$ that I know how to evaluate - the quotient of numbers is the difference of logarithms, so you divide, etc. - but how the heck do you solve one like this with three logs?

I've been able to solve others that only had two logs, but this one confuses me. Help?

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log a - log b - log c = log (a/bc) – Shahab Oct 7 '12 at 17:19
This is an expression which you want to evaluate, not an equation (no equal sign) to solve. – Ross Millikan Oct 7 '12 at 17:49
Thanks for the grammatical contribution. It really helps. – Brandt Oct 7 '12 at 18:03
up vote 1 down vote accepted

We know that $\log(a)- \log(b)= \log(\frac{a}{b})$. Hence, we have

$\log_5(8) - \log_5(20)-\log_5(10) = \log_5(\frac{8}{20})-\log_5(10) $

$= \log_5(\frac{\frac{8}{20}}{10})$

$= \log_5(\frac{8}{200})$

$= \log_5(\frac{1}{25})$

$= \log_5(1)-\log_5(25)$

$= 0 -2 = -2$

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This is exactly how I ultimately solved it. Thanks. :) – Brandt Oct 7 '12 at 23:18

A start: If you know how to handle the sum or difference of two logarithms, you can handle the sum, difference of arbitrarily many. Just deal with them two at a time. For example, $$\log_5 8-\log_5 20-\log_5 10=(\log_5 8 -\log_5 20)-\log_5 10=\log_5(8/20)-\log_5 10=\log_5(?).$$

Added: For a problem like this one, which involves smallish integers, there is another way to proceed. Note that $8=2^3$, so $\log_5 8=3\log_5 2$.

Similarly, $\log_5 20=\log_5(2^2\cdot 5)=2\log_5 2+\log_5 5=2\log_5 2+1$. Similarly, $\log_5 10=\log_5 2+1$. So our expression is equal to $$3\log_5 2-(2\log_5 2+1)-(\log_5 2+1).$$ Simplify. The $\log_5 2$ terms disappear.

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$$\log_5 8-\log_5 20-\log_5 10=\log_5 8/200=\log_5 1/25=\log_5 1-\log_5 25=0-2=-2 $$

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Implement the formula:

$1)$ $\log_x a-\log_x b=\log_x \frac{a}{b}$

$2)$ $\log_a 1=\log_a a^0=0$

$3)$ $\log_x a^n=n\log_x a$

$\log_5 8-\log_5 20-\log_5 10=\log_5\frac{8}{20}-\log_5 10=\log_5\frac{2}{5}-\log_5 10=\log_5\frac{\frac{2}{5}}{10}=\log_5 \frac{2}{50}=\log_5 \frac {1}{25}=\log_5 1 -\log_5 25=0-\log_5 5^2=0-2=-2$

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