${}^2\log \hspace{0.17em}(3)\approx \mathrm{1,6}$; ${}^2\log \hspace{0.17em}(5)\approx \mathrm{2,3}$; ${}^2\log \hspace{0.17em}(15)\approx \mathrm{3,9}$

${}^2\log \hspace{0.17em}(3)+{}^2\log \hspace{0.17em}(5)={}^2\log \hspace{0.17em}(15)$

Zeker fout, want ${}^2\log \hspace{0.17em}(8)=3$

Vermoeden: ${}^3\log \hspace{0.17em}(4)+{}^3\log \hspace{0.17em}(10)={}^3\log \hspace{0.17em}(40)$;
$\mathrm{1,26}+\mathrm{2,10}=\mathrm{3,36}$, dus lijkt te kloppen.

$2^a=3$, $2^b=5$ en $2^c=15$

$3\cdot 5=15$, dus $2^a\cdot 2^b=2^{a+b}=2^c$, dus $a+b=c$

-

$3$ uur; $1$ uur; $3+1=4$ uur

$2$ uur; $\frac{1}{2}$ uur; $2+\frac{1}{2}=2\frac{1}{2}$ uur

In $x+y$ uur

1. ${}^5\text{log}(625)+{}^5\text{log}(\frac{1}{5})$ $=$ $4+\mathrm{‐}1=3$

2. ${}^5\text{log}(625)+{}^5\text{log}(\frac{1}{5})$ $=$ ${}^5\text{log}(125)$ $=3$

$\text{log}(20)+\text{log}(5)\approx \mathrm{1,3}+\mathrm{0,7}=2$
$\text{log}(20)+\text{log}(5)=\text{log}(100)=2$ en
$\text{log}(5)+\text{log}(\frac{1}{2})\approx \mathrm{0,7}+\mathrm{‐}\mathrm{0,3}=\mathrm{0,4}$
$\text{log}(5)+\text{log}(\frac{1}{2})=\text{log}(\mathrm{2,5})\approx \mathrm{0,4}$

In beide gevallen vind je $\mathrm{1,5}$ respectievelijk $\mathrm{3,38039...}$

${}^g\log \hspace{0.17em}(\frac{1}{x})={}^g\log \hspace{0.17em}(x^{\mathrm{‐}1})=\mathrm{‐}1\cdot {}^g\log \hspace{0.17em}(x)=\mathrm{‐}{}^g\text{log}(x)$

$\text{log}(ab)+\text{log}(bc)+\text{log}(ca)=\text{log}(ab\cdot bc\cdot ca)=\text{log}\left({(abc)}^2\right)=2\text{log}(abc)$

$\text{log}\left(\frac{a}{b}\right)+\text{log}\left(\frac{b}{c}\right)+\text{log}\left(\frac{c}{a}\right)=\text{log}\left(\frac{abc}{abc}\right)=\text{log}(1)=0$

• ${}^2\text{log}(x)=3$ $\to$ $x=2^3$ $\to$ $x=8$

• ${}^3\text{log}(x+1)=3$ $\to$ $x+1=3^3=27$ $\to$ $x=26$

• ${}^4\text{log}(2x)=\mathrm{‐}3$ $\to$ $2x=4^{\mathrm{‐}3}=\frac{1}{64}$ $\to$ $x=\frac{1}{128}$

• ${}^5\text{log}(2x+1)=3$ $\to$ $2x+1=5^3=125$ $\to$ $x=62$

• ${}^3\text{log}(\sqrt{x})=2$ $\to$ $\sqrt{x}=3^2=9$ $\to$ $x=81$

• ${}^2\text{log}(x^2-1)=3$ $\to$ $x^2-1=2^3=8$ $\to$ $x^2=9$ $\to$ $x=\mathrm{‐}3$ of $x=3$

• ${}^5\text{log}(\frac{1}{x})=\mathrm{‐}2$ $\to$ $\frac{1}{x}=5^{\mathrm{‐}2}=\frac{1}{25}$ $\to$ $x=25$

• ${}^2\text{log}(x^2-2x)=3$ $\to$ $x^2-2x=2^3=8$ $\to$ $x^2-2x-8=0$ $\to$ $(x-4)(x+2)=0$ $\to$ $x=4$ of $x=\mathrm{‐}2$

• $\log (x)=3\cdot \log (6)$ $\to$ $\log (x)=\log (6^3)=\log (216)$ $\to$ $x=216$

• $3\cdot \log (x)=\log (6)$ $\to$ $\log (x^3)=\log (6)$ $\to$ $x^3=6$ $\to$ $x=\sqrt[3]{6}$

• $2\cdot \log \left(\frac{1}{x}\right)=3\cdot \log (4)$ $\to$ $\log \left({(\frac{1}{x})}^2\right)=\log (4^3)$ $\to$ $\frac{1}{x^2}=64$ $\to$ $x^2=\frac{1}{64}$ $\to$ $x=\mathrm{‐}\frac{1}{8}$ (vervalt!) of $x=\frac{1}{8}$, dus $x=\frac{1}{8}$

• $\log (6)+\log \left(\frac{1}{x}\right)=\log (x)$ $\to$ $\log \left(\frac{6}{x}\right)=\log (x)$ $\to$ $\frac{6}{x}=x$ $\to$ $x^2=6$ $\to$ $x=\mathrm{‐}\sqrt{6}$ (vervalt!), of $x=\sqrt{6}$, dus $x=\sqrt{6}$

• ${}^2\log \hspace{0.17em}(8x)={}^2\log \hspace{0.17em}(12)$ $\to$ $8x=12$ $\to$ $x=1\frac{1}{2}$

• ${}^2\log \hspace{0.17em}\left(\frac{95}{x}\right)={}^2\log \hspace{0.17em}(5)$ $\to$ $\frac{95}{x}=5$ $\to$ $x=19$

• ${}^5\log \hspace{0.17em}\left(\frac{x}{2}\right)={}^5\log \hspace{0.17em}(7)$ $\to$ $\frac{x}{2}=7$ $\to$ $x=14$

• $\log (40x)=4$ $\to$ $40x={10}^4=10000$ $\to$ $x=250$

• $\log \left(\frac{x}{5}\right)=\log (10)+\log (7)=\log (70)$ $\to$ $\frac{x}{5}=70$ $\to$ $x=350$

• ${}^2\log \hspace{0.17em}\left(\frac{x}{3}\right)={}^2\log \hspace{0.17em}\left(\frac{12}{x}\right)$ $\to$ $\frac{x}{3}=\frac{12}{x}$ $\to$ $x^2=36$ $\to$ $x=\mathrm{‐}6$ (vervalt!) of $x=6$, dus $x=6$