# Can this log question be simplified?

${ 2^{log_3 5}} - {5^{log_3 2}}.$ I don't know any formula that can apply to it or is there a formula?? Even a hint will be helpful.

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Use this: $$x^{\log_by}=y^{\log_bx}.$$

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This made it so easy Thanks . IT'S the shortest and the fastest method –  Sudhanshu Aug 10 '14 at 14:13

A symmetrical approach:

$$(\log_n b)(\log_n a)=(\log_n a )(\log_n b)\\ \log_na^{log_n b}=\log_nb^{log_n a}\\ a^{log_n b}=b^{log_n a}$$ Put $a=2, b=5, n=3$:

$$2^{log_3 5}=5^{log_3 2}\\ \Rightarrow 2^{log_3 5}-5^{log_3 2}=0$$

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Let $\displaystyle A=2^{\log_35}\implies \log A=\frac{\log 5}{\log 3}\log 2$

Similarly, $\displaystyle B=5^{\log_32}\implies\log B=\cdots$

$\displaystyle\implies\log B=\log A\implies\cdots$ as $A,B$ are real

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$${ 2^{\log_3 5}} - {5^{\log_3 2}}= \\ 3^{\log_3 { 2^{\log_3 5}} }-3^{\log_3 {5^{\log_3 2}}}= \\ 3^{\log_3 5 \cdot \log_3 { 2} }-3^{\log_3 2 \cdot \log_3 {5}}= \\ \left ( 3^{\log_3 5 }\right )^{\log_3 { 2}} -\left ( 3^{\log_3 2 }\right )^{\log_3 {5} }= \\ 5^{\log_3 2 }-2^{\log_2 5} \Rightarrow \\ \\ { 2^{\log_3 5}} - {5^{\log_3 2}}=5^{\log_3 2 }-2^{\log_2 5} \Rightarrow \\ 2 \cdot 2^{\log_3 5}=2 \cdot 5^{\log_3 2} \Rightarrow \\ \\ \\ 2^{\log_3 5}= 5^{\log_3 2}$$



So we conclude that $${ 2^{\log_3 5}} - {5^{\log_3 2}}= 5^{\log_3 2} - {5^{\log_3 2}}=0$$

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${ 2^{\log_3 5}} - {5^{\log_3 2}}$

$= 2^{\log_3 5} - 5^{\frac{\log_5 2}{\log_5 3}}$ (change of base)

$= 2^{\log_3 5} - (5^{\log_5 2})^{(\frac{1}{\log_5 3})}$ (using $a^{\frac{b}{c}} = {(a^b)}^{\frac{1}{c}}$)

$= 2^{\log_3 5} - 2^{\frac{1}{\log_5 3}}$

$= 2^{\log_3 5} - 2^{\frac{\log_5 5}{\log_5 3}}$ ($\because \log_5 5 = 1$)

$=2^{\log_3 5} - 2^{\log_3 5}$ (basically a "reverse" change of base)

$=0$

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