Proof of the power rule for logarithms

Question

What is the proof for the power rule for logarithms? And are there different ways to prove it?

Answer 1

$$\log_a(M^p)=p\log_a(M)$$

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 $$\log_a(M^p)=\log_a(a^{pt})=pt=p\log_a(M).$$

Answer 2

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!

Answer 3

$$e^{a \cdot \ln(x)}=\left( e^{\ln(x)} \right)^a=x^a$$

$$e^{a \cdot \ln(x)}=x^a$$

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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Answer 4

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?

Answer 5

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.

Answer 6

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 $a^k = a^{λn}$

$$a^k = a^{λn} \implies k = λn \implies k = n\log_a(x)$$ $$\therefore k = n\log_a(x) \implies \log_a(x^n) = n\log_a(x)$$