logarithms properties 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:
$$2=3^{\frac{\ln{2}}{\ln{3}}}$$
 A: Taking the natural log on both sides, $$\ln(2) = \ln(3^{\frac{\ln(2)}{\ln(3)}}) = \frac{\ln(2)}{\ln(3)}\times \ln(3) $$
A: 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.
A: 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.
$$ 2=2. $$
A: 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 $3$.
A: With positive $a$ and $b$, $a^b=e^{b \ln a}$. Hence $3^{\frac{\ln 2}{\ln 3}}=e^{\ln 2}=2$.
