Evaluating a logarithm.

Question

I cant seem to grasp how to evaluate this specific example $\log_5(287)$ as I dont know, what to do in the case of the base not fitting perfectly into the $(x)$ value.

Answer 1

HINT

To simplify recall that

$$\log_a b=\frac{\log_c a}{\log_c b}$$ according to the base $c>0$ and $c\neq 1$ we prefer.

Answer 2

Using some number-theoretic coincidences and the Taylor series $\log(1+x)=x-{1\over2}x^2+\cdots$ when $|x|\lt1$, one can show

$${1\over3}\log_5287={\log287\over3\log5}={\log287\over\log125} ={\log(288-1)\over\log(128-3)} ={\log(256+32)+\log(1-{1\over288})\over\log128+\log(1-{3\over128})}\\ ={8\log2+\log(1+{1\over8})+\log(1-{1\over288})\over7\log2+\log(1-{3\over128})}\\ \approx{8\log2+({1\over8}-{1\over128})-{1\over288}\over7\log2-{3\over128}}$$

where we've truncated the Taylor series as soon as the terms look suitably small. If you're satisfied with a crude approximation, use $\log2\approx0.7$, $1/8=.125$, $1/128\approx0.01$, $1/288\approx1/250=0.004$, and $3/128\approx3/120=1/40=.025$, and get

$${5.6+.125-.01-.004\over4.9-.025}={5.711\over4.875}\approx{5700\over4875}={76\over65}\approx1+{11\over66}\approx1.17$$

It follows that

$$\log_5287\approx3.51$$

which is not a bad approximation to the "exact" value, $3.516434\ldots$.

Answer 3

I thing that you must memorize four logarithms in base $e$, namely those of $\color{red}{2}$, $\color{red}{3}$, $\color{red}{5}$ and $\color{red}{7}$, since $$\log(2)=0.693147$$ $$\log(3)=1.09861$$ $$\log(4)=2\log(2)=1.38629$$ $$\log(5)=1.60944$$ $$\log(6)=\log(2)+\log(3)=1.79176$$ $$\log(7)=1.94591$$ $$\log(8)=3\log(2)=2.07944$$ $$\log(9)=2\log(3)=2.19722$$ $$\log(10)=\log(2)+\log(5)=2.30259$$

So $$\log_5(287)=\frac{\log (287)}{\log (5)}=\frac{\log \left(\frac{287}{300}\right) +\log(300)}{\log (5)}=\frac{\log(3)+2\log(2)+2\log(5)+ \log \left(\frac{287}{300}\right)}{\log (5)}$$

Now, for the last one, consider the very fast convergence expansion $$\log \left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5} \right)+O\left(x^7\right)$$ Let $$\frac{1+x}{1-x}=\frac{287}{300}\implies x=-\frac{13}{587}$$

Using the first term only and the few basic values, you should get $3.51644$ while the exact value is $3.51643$.