If $a^x=b^y$, then how come $x\log a=y\log b$ holds? Can anyone show me how this is with all steps and necessary logarithm formula?
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6$\log x^r=r\log x$. – David Mitra May 21 '15 at 13:24
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@DavidMitra-can you please say what is 'a'? – Soham May 21 '15 at 13:29
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1Had a typo at first; it's the correct formula now. Take logs of both sides of your equation and apply the above formula. – David Mitra May 21 '15 at 13:32
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@DavidMitra-That was why I was not being able to understand. – Soham May 21 '15 at 13:33
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$\log a^x = \log \overbrace{(a\cdot a \cdots \cdots a)}^x = \log a + \log a + \cdots \cdots + \log a = x\log a$
Since $a^x = b^y\quad $ so $\quad \log a^x = \log b^y \implies x\log a = y\log b$
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