Proving that $\log_b(r^s) = s\log_b(r)$

Question

The Question

If $b,r,s \in \mathbb{R^+}$ prove that $\log_b(r^s) = s\log_b(r)$

My Work

1) $\log_b(r^s)$

2) $s$ can be expressed as the sum of an integer part $n$ and a real part $m$: $s = m + n$

3) $\log_b(r^{n+m})$

4) $\log_b(r^nr^m)$

5) $\log_b(r^n) + \log_b(r^m)$

6) $\log_b(r\cdot r\cdot r \cdots r) + \log_b(r^m)$

7) $\log_b(r) + \log_b(r) + \log_b(r) + \cdots + \log_b(r) + \log_b(r^m)$

8) $n\log_b(r) + \log_b(r^m)$

Where I am Having Trouble

I'm having trouble getting the $m$ in front of the second summand which I feel is necessary for the theorem. How can I finish off this proof? I was given this rule as a secondary school student with no proof and would like to have it explained.

Answer 1

Define notations and use the definition of logarithm:

$$\log_br=x\stackrel{\text{By def.}}\iff b^x=r$$

and from here

$$\log_br^s=t\iff \color{red}{b^t}=r^s=(b^x)^s=\color{red}{b^{xs}}\implies t=xs\;\;\;Q.E.D.$$

Answer 2

$$b^{\log_b (r^s)}=r^s=(b^{\log_b(r)})^s=b^{s\log_b(r)},$$ which implies $$b^{\log_b(r^s)-s\log_b(r)}=1,$$ which implies $$\log_b(r^s)-s\log_b(r)=0.$$

Answer 3

Take $\log_b (r^s):=F(s)$ ($r$ an arbitrary constant) and note that $F(0)=\log_b (1)=0$

Show that:

$F(s)=F(s-1)+\log_b r$

Hence $\frac{F(s)-F(s-1)}{s-(s-1)}=\log_b (r)$ and $F$ is of slope $\log_b (r)$:

$$F(s)=(\log_b r)s+c$$

But $F(0)=0$ gives $c=0$.

If we proved it for all positive arbitrary constants $r>0$, then the equation must hold for all $r>0$.

Answer 4

$$\log_b(x) = y \iff b^y=x$$ Using the definition we note that $$b^{s\log_b(r)}=(b^{\log_b(r)})^s=r^s$$ So $\log_b(r^s) = s\log_b(r)$.