- UsukiDoll

logarithm refresher question. So suppose I have... (will draw)

- schrodinger

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- UsukiDoll

|dw:1437812609759:dw|

- UsukiDoll

or
\[\LARGE \log(\frac{a}{b^2})\]
and I want to use log rules to separate them

- UsukiDoll

would the subtraction rule and the exponent rule be used at the same time?

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## More answers

- UsukiDoll

ok yes that's the subtraction rule for that problem.
so now I put the exponent 2 in the front right?

- UsukiDoll

|dw:1437812695740:dw|

- UsukiDoll

alright... I guess I got log rules and elevator rules confused (that's why it got mashed up) so use pure log rules even those the exponent is in the denominator?

- anonymous

Yes what are elevator rules?

- UsukiDoll

oh the elevator rule is for negative exponents. like for example suppose we have
\[\LARGE x^{-2}\]
since we can't have negative exponents we have to write it as
\[\LARGE \frac{1}{x^2}\] as in bring it downstairs

- anonymous

Using this Log Calculator http://www.acalculator.com/logarithm-calculator-logx-logarithmic-equations.html it is easiest to make calculation for logarithms.

- UsukiDoll

now if we have something like
\[\LARGE \frac{1}{x^{-2}}\]
since negative exponents aren't allowed I have to change this.. bring it upstairs
\[\LARGE x^{-(-2)} \rightarrow x^2 \]

- UsukiDoll

@post thanks but I rather do them manually :)

- UsukiDoll

@Deeezzzz why did you delete some of your posts? I wanna give you a medal

- UsukiDoll

ok... let's have another example ummmm
\[\LARGE \log(\frac{a^{6}b^{7}}{c^4})\]

- UsukiDoll

alright so all 3 log rules are being used here.

- UsukiDoll

\[\LARGE \log a^6+ \log b^7 -\log c^4 \] that's the first part

- UsukiDoll

addition and subtraction rules being used.. now for exponents
\[\LARGE 6\log a+ 7\log b -4 \log c\]

- UsukiDoll

ah I get it.. pure log rules ^_^ and log rules only ^_^

- UsukiDoll

cookie monster! :D

- mathmath333

hello

- UsukiDoll

@mathmath333 got any logs similar to what I did so I can practice?

- mathmath333

yea

- UsukiDoll

k.

- mathmath333

\(\Large \color{black}{\begin{align} \log_{2} (9-2^{x})=10^{\log_{10} (3-x)}
\end{align}}\)

- UsukiDoll

not those -_-

- UsukiDoll

the one you had earlier @mathmath333

- mathmath333

only one question was there like that

- mathmath333

any way that is also interesting question

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