Zoe Oneal

2021-11-06

Given that ${\mathrm{log}}_{a}\left(5\right)\approx 0.65$ and ${\mathrm{log}}_{a}\left(3\right)\approx 0.44$, evaluate each of the following. Hint: use the properties of logarithms to rewrite the given logarithm in terms of the logarithms of 5 and 3
a)${\mathrm{log}}_{a}\left(0.6\right)$
b)${\mathrm{log}}_{a}\left(\sqrt{3}\right)$
c)${\mathrm{log}}_{a}\left(15\right)$
d)${\mathrm{log}}_{a}\left(25\right)$
e)${\mathrm{log}}_{a}\left(75\right)$
f)${\mathrm{log}}_{a}\left(1.8\right)$

### Answer & Explanation

jlo2niT

Step1
${\mathrm{log}}_{a}\left(5\right)\approx 0.65{\mathrm{log}}_{a}\left(3\right)\approx 0.44$
properties of logarithms are
1.$\mathrm{log}\left(ab\right)={\mathrm{log}}_{c}\left(a\right)+{\mathrm{log}}_{c}\left(b\right)$
2.${\mathrm{log}}_{c}\left(\frac{a}{b}\right)={\mathrm{log}}_{c}\left(a\right)-{\mathrm{log}}_{c}\left(b\right)$
3.${\mathrm{log}}_{c}\left({a}^{b}\right)=b{\mathrm{log}}_{c}a$
Step2
${\mathrm{log}}_{a}\left(0.6\right)={\mathrm{log}}_{a}\left(\frac{3}{5}\right)$
${\mathrm{log}}_{a}\left(0.6\right)={\mathrm{log}}_{a}\left(3\right)-{\mathrm{log}}_{a}\left(5\right)$
${\mathrm{log}}_{a}\left(0.6\right)=0.44-0.65$
${\mathrm{log}}_{a}\left(0.6\right)=-0.21$
Step3
${\mathrm{log}}_{a}\left(\sqrt{3}\right)={\mathrm{log}}_{a}\left({3}^{\frac{1}{2}}\right)$
${\mathrm{log}}_{a}\left(\sqrt{3}\right)=\frac{1}{2}{\mathrm{log}}_{a}\left(3\right)$
${\mathrm{log}}_{a}\left(\sqrt{3}\right)=\frac{1}{2}\cdot 0.44$
${\mathrm{log}}_{a}\left(\sqrt{3}\right)=0.22$
Step4
${\mathrm{log}}_{a}\left(15\right)={\mathrm{log}}_{a}\left(5\cdot 3\right)$
${\mathrm{log}}_{a}\left(15\right)={\mathrm{log}}_{a}\left(5\right)+{\mathrm{log}}_{a}\left(3\right)$
${\mathrm{log}}_{a}\left(15\right)=0.65+0.44$
${\mathrm{log}}_{a}\left(15\right)=1.09$
Step5
${\mathrm{log}}_{a}\left(25\right)={\mathrm{log}}_{a}\left({5}^{2}\right)$
${\mathrm{log}}_{a}\left(25\right)=2{\mathrm{log}}_{a}\left(5\right)$
${\mathrm{log}}_{a}\left(25\right)=2\cdot 0.65$
${\mathrm{log}}_{a}\left(25\right)=1.3$
Step6
${\mathrm{log}}_{a}\left(75\right)={\mathrm{log}}_{a}\left(25\cdot 3\right)$

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