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I've been looking into the logarithm change of base rule lately. I get the proof, but I feel I would understand it deeper and more intuitively if I also understood what was going on when we changed base from the perspective of exponents. After all, all the other rules of logarithms can be derived from the laws of exponents, and make more sense once we do so. So can someone please show me, how the change of base rule can be written in form of exponents? What are we doing, from the perspective of exponents, when we are using the change of base rule?

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It derives from if $a^k = b$ and $b^m = w$ then $ a^{km}=(a^k)^m = b^m = w$.

So $\log_a b = k$ and $\log_b w = m$ then $\log_a w =k*m= \log_a b * \log_b w$

So $\log_b w = \frac {\log_a w}{\log_b w}$

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Suppose $a^n=b^m=w $

$b=a^{log_a b} $ by definition.

So $w=b^m=(a^{\log_a b})^m=a^{m\log_a b}=a^n $ so

$n=\log_a w= m\log_a b =\log_b w\log_a b $. (Assuming logs are injective.)

Intuitively for $1<a <b $ we must raise $a $ to a higher power than $b$ to get the same result. How high a power? Well, $\log_a b $ as $a^{k*\log_a b}=b^k $.

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Recall that by definition for $a\neq 1$ and $b>0$ we have that

$$c=\log_a b \iff a^c=b$$

and since in base $x\neq 1$ we have

  • $a=x^{\log_x a}$

  • $b=x^{\log_x b}$

and therefore

$$a^c=b\implies (x^{\log_x a})^c=x^{\log_x b}\implies x^{c\log_x a}=x^{\log_x b}\implies c\log_x a=\log_x b$$

that is

$$c=\log_a b=\frac{\log_x b}{\log_x a}$$

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For positives $a$, $b$ and $c$, $c\neq1$ and $b\neq1$ we obtain: $$c^{\log_ca}=a=b^{\log_ba}=\left(c^{\log_cb}\right)^{\log_ba}=c^{\log_ab\log_cb},$$ which gives $$\log_ba=\frac{\log_ca}{\log_cb}$$

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In terms of exponents, the change of base rule is equivalent to the statement for positive numbers $b,c,r,s$ given $b\ne1$ and $c\ne1$:

$$\text{If }b=c^r\text{ then }b^s=c^{rs} $$

Let $x=b^s$ and $x=c^t$ and $b=c^r$. Then we have both that $x=(c^r)^s=c^{rs}$ and that $x=c^t$. So it must be the case that $rs=t$.

Now let us write $r,s,t$ in terms of logarithms.

$$x=b^s\text{ is equivalent to }s=\log_bx $$ $$x=c^t\text{ is equivalent to }t=\log_cx $$ $$b=c^r\text{ is equivalent to }r=\log_cb $$

Therefore, the conclusion above that $rs=t$ is equivalent to the statement

$$ \log_cb\cdot\log_bx=\log_cx $$ from which we get

$$ \log_bx=\frac{\log_cx}{\log_cb} $$

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