What is the proof for the power rule for logarithms? And are there different ways to prove it?
$\endgroup$ 06 Answers
$\begingroup$$$\log_a(M^p)=p\log_a(M)$$
Proof : Let $t=\log_a(M)$. Then, by definition, we have $a^t=M$. So, we have $$M^p=(a^t)^p=a^{pt}.$$ Hence, we have $$\log_a(M^p)=\log_a(a^{pt})=pt=p\log_a(M).$$
$\endgroup$ 2 $\begingroup$Here is one way to show the identity. Define the natural logarithm as
$$\log x=\int_1^x \frac{dt}{t}$$
for $x>0$. Then, we have
$$\log x^n=\int_1^{x^n} \frac{dt}{t}=\sum_{k=1}^n\int_{x^{k-1}}^{x^k}\frac{dt}{t} \tag 1$$
Now substituting $t=x^{k-1}u$ in $(1)$ reveals that
$$\begin{align} \log x^n&=\sum_{k=1}^n\int_{1}^{x}\frac{du}{u}\\\\ &=n\log x \end{align}$$
as was to be shown!
$\endgroup$ 2 $\begingroup$$$e^{a \cdot \ln(x)}=\left( e^{\ln(x)} \right)^a=x^a$$
$$e^{a \cdot \ln(x)}=x^a$$
The logarithm of the left part can be done easily using the definition.
$$\ln \left( e^{a \cdot \ln(x)} \right)=a \cdot \ln(x)$$
Thus,
$$\ln(x^a)=a\cdot \ln(x)$$
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$\endgroup$ $\begingroup$It depends on how you define the logarithm function. If you define it as the inverse function of the exponential function, then this isn't hard to prove.
Let $x\in\mathbb{R}^+$ and let $a\in\mathbb{R}$.
We want to show that $\ln(x^a)=a\ln(x)$. So first let $y=\ln(x^a)$. What is $\exp(y)=e^y$? Then calculate $e^{a\ln(x)}$. Do you find the same result?
$\endgroup$ $\begingroup$Hint.
If you define $$\ln x =\int_1^x \frac{dt}{t}$$ you can make an integration by substitution to prove the identity you mention.
$\endgroup$ $\begingroup$My approach:
$$\log_a(x^n)=n\log_a(x)$$
Proof : Let $λ=\log_a(x)$, thus $a^λ=x$
$$\log_a(x^n)=k$$ Hence we have $a^k = x^n$
Using $a^λ=x$, we have $a^k = a^{λn}$
$$a^k = a^{λn} \implies k = λn \implies k = n\log_a(x)$$$$\therefore k = n\log_a(x) \implies \log_a(x^n) = n\log_a(x)$$
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