$x = 2^{\log_57}$ and $y = 7^{\log_52}$ find $x - y$

Solution 1:

You can show that $a^{\log_bc}=c^{\log_ba}$ for any positive $a$, $b$ and $c$. Indeed, $$a^{\log_b c}=\left(b^{\log_ba}\right)^{\log_b c}=\left(b^{\log_bc}\right)^{\log_b a}=c^{\log_ba}.$$

Solution 2:

$x = 2^{log_5(7)} = 5^{log_5(2^{log_5(7)})} = 5^{log_5(7)log_5(2)}$

$ = 5^{log_5(2)log_5(7)} = 5^{log_5(7^{log_5(2)})} $

$ = 7^{log_5(2)} = y$

Another method

$x = 2^{log_5(7)} = 2^{\frac{ln(7)}{ln(5)}} = e^{ln(2^{(\frac{ln(7)}{ln(5)})})}$

$ = e^{{\frac{ln(2)ln(7)}{ln(5)}}} = e^{ln(7^{(\frac{ln(2)}{ln(5)})})}$

$ = 7^{\frac{ln(2)}{ln(5)}} = 7^{log_5(2)} = y$

Hence, $x = y$

$x-y = 0$

Solution 3:

$\large x=2^\frac{\ln 7}{\ln 5}\qquad → \ln(x) = \left(\frac{\ln 7}{\ln 5}\right) (\ln 2)$
$\large y=7^\frac{\ln 2}{\ln 5}\qquad → \ln(y) = \left(\frac{\ln 2}{\ln 5}\right) (\ln 7)$

(x,y) have same log value, thus $\;x-y = x-x = 0$