Wolfram Alpha says the answer is zero. But how to find the value of $x - y$ when the bases aren't even same. I know that $K^{\log_K x}= x.$
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2Hint: Try showing $x=y$ by proving $\log_5(x) = \log_5(y)$. – Andreas Tsevas Dec 02 '21 at 15:42
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1@Andreas Gotcha, both x and y are indeed same in my problem. Each being [log7 * log 2] / log 5. Thanks. – Emanat S Dec 02 '21 at 15:51
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What is $\log[x,y]?$ Is it $\log_xy$ or $\log_yx,$ or something else? – Thomas Andrews Dec 02 '21 at 16:11
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Hint: $$\log_b c=\frac{\log_ac}{\log_ab}.$$ – Thomas Andrews Dec 02 '21 at 16:14
3 Answers
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$x = 2^{log_5(7)} = 5^{log_5(2^{log_5(7)})} = 5^{log_5(7)log_5(2)}$
$ = 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, $x = y$
$x-y = 0$
User 123
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You can show that $a^{\log_bc}=c^{\log_ba}$ for any positive $a$, $b$ and $c$. Indeed, $$a^{\log_b c}=\left(b^{\log_ba}\right)^{\log_b c}=\left(b^{\log_bc}\right)^{\log_b a}=c^{\log_ba}.$$
Zuy
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$\large x=2^\frac{\ln 7}{\ln 5}\qquad → \ln(x) = \left(\frac{\ln 7}{\ln 5}\right) (\ln 2)$
$\large y=7^\frac{\ln 2}{\ln 5}\qquad → \ln(y) = \left(\frac{\ln 2}{\ln 5}\right) (\ln 7)$
(x,y) have same log value, thus $\;x-y = x-x = 0$
albert chan
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