Help understanding $a^w=1+kw$

Question

I have a question.

Euler starts with this situation: $$a>1$$ $$a^w=1+kw$$

His example was this:

Let $a = 10$ , we look for the logarithm of a number which exceeds $1$ by the smallest possible amount, for instance: $$1+\frac{1}{1000000}$$ so that $$kw=\frac{1}{1000000}$$ Then:$$\log{(1+\frac{1}{1000000})}=\log{(\frac{1000001}{1000000})}=0.00000043429=w.$$

$k$ is then: $$k=2.30258$$

I don't understand what this equation $a^w=1+kw$ does mean.

Answer 1

The equation $a^w = 1+kw$ isn't actually valid. Rather you have $a^w = 1+kw+O(w^2)$ i.e. $a^w \approx 1+kw$ when $w$ is small. But for small $w$ the nonlinear term ($O(w^2)$) will be so small that it in practice can be neglected.

EXPLANATION

Using Mac-Laurin Series $a^w=1+wa^0 \ln(a)+\frac{w^2}{2!}a^0(\ln(a))^2+\cdots=1+w\ln(a)+O(w^2)=1+kw+O(w^2), \ k=\ln(a).$

Answer 2

Thanks for demonstrating this idea in the example above in which you're estimating w.

I will explain the meaning of $a^w=1+kw$, given that $a>1$.

From Abstract Algebra, $a>b$ mean $\exists \ k >0$ such that $a=b+k$

If $b=1$, then $a=1+k$ From binomial theorem, you can show that $a^w=(1+k)^w \ge 1+kw$ if $w\ge1$. So you can choose $a^w=1+kw$

Similarly if $0<w<1$, you can show that for $a>0$, $a^w \le w(a-1)+1=kw+1, \ k=a-1 $