Can the power property of logarithms be derived from the power property of exponents using the equation

Can the power property of logarithms be derived from the power property of exponents using the equation
If not, explain why. If so, show the derivation.

The Correct Answer and Explanation is:

Yes, the power property of logarithms can be derived from the power property of exponents using the definition of a logarithm.

Power Property of Logarithms:

log⁡b(xr)=r⋅log⁡b(x)\log_b(x^r) = r \cdot \log_b(x)logb​(xr)=r⋅logb​(x)

Derivation:

Start with the definition of a logarithm:log⁡b(x)=ymeansby=x\log_b(x) = y \quad \text{means} \quad b^y = xlogb​(x)=ymeansby=x

Let us apply this to the expression log⁡b(xr)\log_b(x^r)logb​(xr). Suppose:log⁡b(x)=y⇒by=x\log_b(x) = y \quad \Rightarrow \quad b^y = xlogb​(x)=y⇒by=x

Raise both sides to the power of rrr:(by)r=xr(b^y)^r = x^r(by)r=xr

Using the power property of exponents:byr=xrb^{yr} = x^rbyr=xr

Now take the logarithm base bbb of both sides:log⁡b(xr)=log⁡b(byr)\log_b(x^r) = \log_b(b^{yr})logb​(xr)=logb​(byr)

Since log⁡b(bk)=k\log_b(b^k) = klogb​(bk)=k for any kkk:log⁡b(byr)=yr\log_b(b^{yr}) = yrlogb​(byr)=yr

Thus:log⁡b(xr)=r⋅y\log_b(x^r) = r \cdot ylogb​(xr)=r⋅y

But from earlier, we defined log⁡b(x)=y\log_b(x) = ylogb​(x)=y. Therefore:log⁡b(xr)=r⋅log⁡b(x)\log_b(x^r) = r \cdot \log_b(x)logb​(xr)=r⋅logb​(x)

This completes the derivation.

Explanation (300 words):

The power property of logarithms states that the logarithm of a power of a number is equal to the exponent times the logarithm of the base value. This rule can be derived from the properties of exponents and the definition of logarithms. A logarithm essentially asks the question: to what exponent must we raise the base in order to get a certain number? For example, log⁡b(x)=y\log_b(x) = ylogb​(x)=y means that raising bbb to the power yyy gives xxx, or by=xb^y = xby=x.

To understand how this leads to the power property, assume we want to evaluate log⁡b(xr)\log_b(x^r)logb​(xr). From the definition, if by=xb^y = xby=x, then raising both sides to the power rrr gives (by)r=xr(b^y)^r = x^r(by)r=xr. Applying the power rule for exponents, we simplify the left side to byrb^{yr}byr. That means xr=byrx^r = b^{yr}xr=byr, and taking the logarithm of both sides gives log⁡b(xr)=log⁡b(byr)\log_b(x^r) = \log_b(b^{yr})logb​(xr)=logb​(byr). Since the logarithm and exponential functions are inverses, log⁡b(byr)=yr\log_b(b^{yr}) = yrlogb​(byr)=yr. Therefore, log⁡b(xr)=r⋅log⁡b(x)\log_b(x^r) = r \cdot \log_b(x)logb​(xr)=r⋅logb​(x).

This derivation shows a clear link between exponent properties and logarithmic identities. It also highlights the inverse relationship between logarithms and exponents. Understanding this connection helps reinforce why logarithmic properties work as they do and allows for a deeper grasp of algebraic manipulation involving logs and exponents.