Question

If $lo{g}_{10}8=0.90$ find the value of :
(i) $lo{g}_{10}4$
(ii) $log\sqrt{32}$
(iii) $log0.125$

Answer 1

Given:- ${\log }_{10}8=0.90$

To find:-

(i) ${\log }_{10}4$

(ii) ${\log }_{10}\sqrt{32}$

(iii) ${\log }_{10}0.125$

${\log }_{10}8=0.90\left(\text{Given}\right)$

${\log }_{10}\left(2^3\right)=0.9$

$\Rightarrow 3{\log }_{10}2=0.9$

$\Rightarrow {\log }_{10}2=\frac{0.9}{3}=0.3$

(i) ${\log }_{10}4$

$={\log }_{10}\left(2^2\right)$

$=2{\log }_{10}2$

$=2\times 0.3=0.6$

(ii) ${\log }_{10}\sqrt{32}$

$={\log }_{10}\left({\left(32\right)}^{\frac{1}{2}}\right)$

$=\frac{1}{2}{\log }_{10}\left(2^5\right)$

$=\frac{5}{2}{\log }_{10}2$

$=\frac{5}{2}\times 0.3$

$=0.75$

(iii) ${\log }_{10}0.125$

$={\log }_{10}\left({\left(0.5\right)}^3\right)$

$=3{\log }_{10}\left(\frac{5}{10}\right)$

$=3({\log }_{10}5-{\log }_{10}10)$

$=3({\log }_{10}5-{\log }_{10}(2\times 5))$

$=3({\log }_{10}5-({\log }_{10}2+{\log }_{10}5))$

$=3(-{\log }_{10}2)$

$=3\times (-0.3)$

$=-0.9$