Voer in Y1=2^X en Y2=20, stel het venster goed in (bijvoorbeeld $[0,10]\times [0,25]$.
Je vindt met behulp van de Intersect-optie het getal $\mathrm{4,322}$.

Zie de Uitleg .

Na ${}^{2}log\left(100\right)$ uur, dat is ongeveer $\mathrm{6,64}$ uur en dat is $6$ uur en $39$ minuten.

$3000\cdot {\mathrm{0,98}}^t=2800$

${\mathrm{0,98}}^t=\frac{2800}{3000}=\frac{14}{15}$

$t={}^{\mathrm{0,98}}log\left(\frac{14}{15}\right)\approx \mathrm{3,415}$

$x={}^{2}log\left(7\right)\approx \mathrm{2,807}$

$x={}^{3}log\left(81\right)=4$

$x={}^{\frac{1}{3}}log\left(9\right)=-2$

$x={}^{\frac{1}{3}}log\left(\mathrm{0,01}\right)\approx \mathrm{4,192}$

$x={}^{10}log\left(1000000\right)=6$

$x={}^{10}log\left(\mathrm{0,001}\right)=-3$

$x={}^{\mathrm{0,001}}log\left(\mathrm{0,1}\right)=\frac{1}{3}$

$x={}^{\mathrm{0,001}}log\left(100\right)=\mathrm{-}\frac{2}{3}$

${}^{5}log\left(5^3\right)=3$

${}^{5}log\left(5^{-2}\right)=-2$

${}^{4}log\left(4^3\right)=3$

${}^{\frac{1}{4}}log\left({\left(\frac{1}{4}\right)}^{-3}\right)=-3$

${}^{\frac{1}{3}}log\left({\left(\frac{1}{3}\right)}^4\right)=4$

${}^{2}log\left(2^{\frac{1}{2}}\right)=\frac{1}{2}$

$5^3=125$ en $5^4=625$, dus .

${10}^2=100$ en ${10}^3=1000$, dus .

$2^5=32$ en $2^6=64$, dus .

$2^{-3}=\frac{1}{8}$ en $2^{-2}=\frac{1}{4}$, dus .

${\left(\frac{1}{2}\right)}^{-4}=16$ en ${\left(\frac{1}{2}\right)}^{-5)}=32$, dus .

${\left(\frac{1}{3}\right)}^1=\frac{1}{3}$ en ${\left(\frac{1}{3}\right)}^2=\frac{1}{9}$, dus .

$\mathrm{3,1}$

$\mathrm{2,9}$

$\mathrm{5,9}$

$\mathrm{-2,8}$

$\mathrm{-4,3}$

$\mathrm{1,5}$

$3^x=600$, dus $x={}^{3}log\left(600\right)\approx \mathrm{5,8}$.

${\mathrm{1,7}}^t=525$, dus $x={}^{\mathrm{1,7}}log\left(525\right)\approx \mathrm{11,8}$.

${\mathrm{0,6}}^t=\frac{30}{572}$ , dus $x={}^{\mathrm{0,6}}log\left(\frac{30}{572}\right)\approx \mathrm{5,8}$.

${}^{4}log\left(64\right)=3$

${}^{4}log\left(400\right)\approx \mathrm{4,3}$ (met de GR)

${}^{\frac{1}{3}}log\left(60\right)\approx \mathrm{3,7}$ (met de GR)

${}^{\frac{1}{3}}log\left(81\right)=-4$

${}^{\frac{1}{3}}log\left(\frac{1}{81}\right)=4$

${}^{\mathrm{0,1}}log\left(1000000\right)=-6$

$6^1=6$ en $6^2=36$, dus .

$3^3=27$ en $3^4=81$, dus .

${\left(\frac{1}{2}\right)}^{-3}=8$ en ${\left(\frac{1}{2}\right)}^{-4}=16$, dus .

${\left(\frac{1}{3}\right)}^4\approx \mathrm{0,012}$ en ${\left(\frac{1}{3}\right)}^5\approx \mathrm{0,004}$, dus .

$\mathrm{5,026}$

$\mathrm{-8,399}$

$\mathrm{-3,597}$

$\mathrm{0,306}$

${10}^x=\mathrm{0,01}$, dus $x={}^{10}log\left(\mathrm{0,01}\right)=-2$.

$2^x=60$, dus $x={}^{2}log\left(60\right)\approx \mathrm{5,9}$.

${\mathrm{0,8}}^t=\mathrm{0,5}$, dus $t={}^{\mathrm{0,8}}log\left(\mathrm{0,5}\right)\approx \mathrm{3,1}$.

${}^{2}log\left(2^{\mathrm{2,5}}\right)=\mathrm{2,5}$

${}^{\frac{1}{3}}log\left({\left(\frac{1}{3}\right)}^{-3}\right)=-3$

$2^9=512$ en $2^{\left(10\right)}=1024$, dus .
${}^{2}log\left(513\right)=\mathrm{9,003}$.

${\mathrm{0,4}}^{-4}\approx 29$ en ${\mathrm{0,4}}^{-3}=\mathrm{15,625}$, dus .
${}^{\mathrm{0,4}}log\left(25\right)\approx \mathrm{-3,513}$.

$4^x=\frac{35}{6}$, dus $x={}^{4}log\left(\frac{35}{6}\right)\approx \mathrm{1,27}$.

${\mathrm{1,08}}^t=\frac{12}{7}$, dus $t={}^{\mathrm{1,08}}log\left(\frac{12}{7}\right)\approx \mathrm{7,00}$.