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If $p=\lg5$ and $q=\log_{3}2$, express $\log_{3}5$ in terms of $p$ and $q$.

Um really confused! How do I solve this?

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  • $\begingroup$ By 'lg' do you mean the log to base 2 (the usual meaning of that expression)? It would help to write it out explicitly, if so. $\endgroup$ – Steven Stadnicki Feb 19 '14 at 19:54
  • $\begingroup$ @StevenStadnicki I don't know, I just wrote in the way it is written in my textbook. :) $\endgroup$ – Kiara Feb 19 '14 at 19:55
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Hints:

$$\log_35=\frac{\lg5}{\lg3}$$

$$\lg3=\frac1{\log_32}$$

Note: we use above the "change of base" property of logarithms:

$$\log_ax=\frac{\log_bx}{\log_ba}\;,\;\;a,b,x>0\;,\;\;a,b\neq1$$

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We have $$ \frac{1}{p}=\frac{1}{\log 5}=\frac{\log_5 10}{1}=\log_5 2+\log_5 5=1+\log_5 2. $$ Thus $$ \frac{1}{p}-1=\log_5 2, $$ and hence $$ =\frac{p}{1-p}=\frac{1}{\frac{1}{p}-1}=\log_2 5 $$ Finally $$ \log_3 5=\log_3 2 \cdot \log_2 5=q\cdot\frac{p}{1-p}=\frac{pq}{1-p}. $$

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$q=\log_{3}2$

$p=\log_{10}5$

implies $1/p=\log_{5}10$=$1+\log_{5}2$

therefore $\log_{3}5=qp/(1-p)$.

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Try this:

$log_{10}5 = \frac{log_35}{log_310}$

Let $x = log_35$ and simplify $log_310$ into $log_35+log_32$ Then, $p = \frac{x}{x+q}$

Solving for $x$ gives you $\frac{pq}{1-p}$.

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$log_3(5)=\frac{lg5}{lg3}=\frac{lg2}{lg3}\frac{lg5}{lg2}=log_3(2)\frac{lg5}{lg2}=log_3(2)\frac{lg(5)}{1-lg5}=q\frac{p}{1-p}$

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