Logarithm
1.Which of the following statements is not correct?

Explanation:

Solution:

2.If log 2 = 0.3010 and log 3 = 0.4771, the value of log5 512 is:

Explanation:

Solution:
$\begin{array}{rl}& {\mathrm{log}}_{5}512\\ & =\frac{\mathrm{log}512}{\mathrm{log}5}\\ & =\frac{\mathrm{log}{2}^{9}}{\mathrm{log}\left(\frac{10}{2}\right)}\\ & =\frac{9\mathrm{log}2}{\mathrm{log}10-\mathrm{log}2}\\ & =\frac{\left(9×0.3010\right)}{1-0.3010}\\ & =\frac{2.709}{0.699}\\ & =\frac{2709}{699}\\ & =3.876\end{array}$

3.
$\frac{\mathrm{log}\sqrt{8}}{\mathrm{log}8}$
is equal to:

Explanation:

Solution:
$\begin{array}{rl}& \frac{\mathrm{log}\sqrt{8}}{\mathrm{log}8}={\frac{\mathrm{log}\left(8\right)}{\mathrm{log}8}}^{\frac{1}{2}}\\ & =\frac{\frac{1}{2}\mathrm{log}8}{\mathrm{log}8}\\ & =\frac{1}{2}\end{array}$

4.
$\mathrm{I}\mathrm{f}\mathrm{log}\frac{a}{b}+\mathrm{log}\frac{b}{a}=\mathrm{log}\left(a+b\right),\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n}:$

Explanation:

Solution:
$\begin{array}{rl}& \mathrm{log}\frac{a}{b}+\mathrm{log}\frac{b}{a}=\mathrm{log}\left(a+b\right)\\ & ⇒\mathrm{log}\left(a+b\right)=\mathrm{log}\left(\frac{a}{b}×\frac{b}{a}\right)=\mathrm{log}1\\ & So,a+b=1\end{array}$

5.
$\mathrm{I}\mathrm{f}\phantom{\rule{1pt}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{1pt}{0ex}}{\mathrm{log}}_{10}7=a,\phantom{\rule{1pt}{0ex}}\phantom{\rule{1pt}{0ex}}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{1pt}{0ex}}\phantom{\rule{1pt}{0ex}}{\mathrm{log}}_{10}\left(\frac{1}{70}\right)\mathrm{i}\mathrm{s}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{1pt}{0ex}}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{l}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{1pt}{0ex}}\mathrm{t}\mathrm{o}$

Explanation:

Solution:
$\begin{array}{rl}& {\mathrm{log}}_{10}\left(\frac{1}{70}\right)\\ & ={\mathrm{log}}_{10}1-{\mathrm{log}}_{10}70\\ & =-{\mathrm{log}}_{10}\left(7×10\right)\\ & =-\left({\mathrm{log}}_{10}7+{\mathrm{log}}_{10}10\right)\\ & =-\left(a+1\right)\end{array}$

6.If log10 2 = 0.3010, then log2 10 is equal to:

Explanation:

Solution:
$\begin{array}{rl}& {\mathrm{log}}_{2}10\\ & =\frac{1}{{\mathrm{log}}_{10}2}\\ & =\frac{1}{0.3010}\\ & =\frac{10000}{3010}\\ & =\frac{1000}{301}\end{array}$

7.If log10 2 = 0.3010, the value of log10 80 is:

Explanation:

Solution:
$\begin{array}{rl}& {\mathrm{log}}_{10}80\\ & ={\mathrm{log}}_{10}\left(8×10\right)\\ & ={\mathrm{log}}_{10}8+{\mathrm{log}}_{10}10\\ & ={\mathrm{log}}_{10}\left({2}^{3}\right)+1\\ & =3{\mathrm{log}}_{10}2+1\\ & =\left(3×0.3010\right)+1\\ & 1.9030\end{array}$

8.
$\begin{array}{rl}& \text{The}\phantom{\rule{1pt}{0ex}}\text{value}\phantom{\rule{1pt}{0ex}}\text{of}\\ & \left(\frac{1}{{\mathrm{log}}_{3}60}+\frac{1}{{\mathrm{log}}_{4}60}+\frac{1}{{\mathrm{log}}_{5}60}\right)\text{is}:\end{array}$

Explanation:

Solution:
Given expression
$\begin{array}{rl}& ={\mathrm{log}}_{60}3+{\mathrm{log}}_{60}4+{\mathrm{log}}_{60}5\\ & ={\mathrm{log}}_{60}\left(3×4×5\right)\\ & ={\mathrm{log}}_{60}60\\ & =1\end{array}$

9.If log 2 = 0.30103, the number of digits in 264 is:

Explanation:

Solution:
$\begin{array}{rl}& \mathrm{log}\left({2}^{64}\right)\\ & =64×\mathrm{log}2\\ & =\left(64×0.30103\right)\\ & =19.26592\end{array}$
Its characteristic is 19.
Hence, then number of digits in 264 is 20.

10.
$\mathrm{I}\mathrm{f}\phantom{\rule{thinmathspace}{0ex}}{\mathrm{log}}_{x}\left(\frac{9}{16}\right)=-\frac{1}{2},\phantom{\rule{thinmathspace}{0ex}}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n}\phantom{\rule{thinmathspace}{0ex}}x\phantom{\rule{thinmathspace}{0ex}}\mathrm{i}\mathrm{s}\phantom{\rule{thinmathspace}{0ex}}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{l}\phantom{\rule{thinmathspace}{0ex}}\mathrm{t}\mathrm{o}:$

$\begin{array}{rl}& {\mathrm{log}}_{x}\left(\frac{9}{16}\right)=-\frac{1}{2}\\ & ⇒{x}^{-\frac{1}{2}}=\frac{9}{16}\\ & ⇒\frac{1}{\sqrt{x}}=\frac{9}{16}\\ & ⇒\sqrt{x}=\frac{16}{9}\\ & ⇒x={\left(\frac{16}{9}\right)}^{2}\\ & ⇒x=\frac{256}{81}\end{array}$