Suppose I have the inequality $(\frac{A}{B})^X < (\frac{C}{D})\cdot(\frac{E}{F})^Y$ and I want X by itself.
Can I do this $X\cdot \log(\frac{A}{B}) < \log(\frac{C}{D})\cdot(Y\cdot \log(\frac{E}{F}))$? Am I breaking any rules on the right-hand side?
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Log of a product is the sum of the logs. So you are missing a $+$ sign on the right. – ancient mathematician Sep 30 '20 at 15:34
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Please learn how to use basic Latex before posting a question. Also yes, you are breaking the rules. You need a + sign. – Peanut Sep 30 '20 at 15:34
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3Thanks, I will try to learn latex later on – David Zheng Sep 30 '20 at 15:36
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1$\log (ab)=\log a + \log b$ – KingLogic Sep 30 '20 at 15:45
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It should be $$X\log \left(\frac{A}{B}\right)=\log \left(\frac{C}{D}\right)+Y\log \left(\frac{E}{F}\right)$$ You did not break the log into addition as $\log xy =\log x +\log y$
rash
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Yes!!! You need to apply the rule of addition for logarithmic functions.
Right hand side should be log(C / D) + (Y * log(E / F)).
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One of the logarithm rules is $\log (ab)=\log a + \log b$. Therefore, on the right, it should be $\log (\frac{C}{D}) + Y \log (\frac{E}{F})$.
KingLogic
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