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Theorems on Integration

If log a m...

Question

If ${log}_{a} m = n,$ express $a^{n - 1}$ in terms of $a$ and $m .$

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Solution

${log}_{a} m = n$
So $m = a^{n}$
$\frac{m}{a} = \frac{a^{n}}{a} = a^{n - 1}$
Hence $a^{n - 1} = \frac{m}{a}$

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