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I know it's very easy and naive, but apparently I cannot understand the following equation. Can you please prove it? Thanks in advance.

The equation is:

$$2=3^{\frac{\ln{2}}{\ln{3}}}$$

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    $\begingroup$ Try taking a natural log on both sides and see what happens. $\endgroup$ – user296602 Jan 11 '16 at 19:25
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Taking the natural log on both sides, $$\ln(2) = \ln(3^{\frac{\ln(2)}{\ln(3)}}) = \frac{\ln(2)}{\ln(3)}\times \ln(3) $$

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Starting with: $$2=3^{\frac{\ln{2}}{\ln{3}}}$$

Use the identity that $\frac{\ln2}{\ln3} = \log_3(2)$

Then,

$$2=3^{\frac{\ln{2}}{\ln{3}}} \rightarrow 2 = 3^{\log_3(2)}$$

So taking the $\log_3$ of both sides and bringing the exponent on RHS down,

$$\log_3(2) = \log_3(3^{\log_3(2)}) \rightarrow \log_3 (2) =\log_3(2) *\log_3(3) $$

Now, knowing that $\log_3(3) =1$ we get,

$$\log_3(2) = \log_3(2) $$

Which is we know is true.

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You need to appreciate that log and exponentiation run in opposite ways. ( like + and -)

$$2=3^{\dfrac{\log{2}}{\log{3}}}$$

But as $$\dfrac{\ln2}{\ln3} = \log_3(2),$$

So

$$2=3^ {\ log_3 2 } $$

Since you are doing an operation and its anti operation ( log taking with respect to 3 as base and exponentiation with respect to same base 3 ), 2 is left as it is, as 2 itself.

$$ 2=2. $$

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This is just an application of the change of base formula for logarithms.

$$\frac {\ln(2)} {\ln(3)} = \log_3(2)$$

So we're changing from base $e$ to base $3$.

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With positive $a$ and $b$, $a^b=e^{b \ln a}$. Hence $3^{\frac{\ln 2}{\ln 3}}=e^{\ln 2}=2$.

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