Question

The value of $\text {log}_{0.008}\sqrt{5}+\text{log}_{\sqrt{3}}81-7$ is equal to

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{5}{6}$
• D. $\frac{7}{6}$

Answer 1

Answer: (C)

$\frac{5}{6}$

Answer 2

The given expression can be simplified as follows:

$log_{0.008}\sqrt{​5}​+log_{\sqrt{3}}\ ​​81−7$
For $log⁡_{0.008}\sqrt{5}​:$

$=log_{5^{−3}}​(5^{\frac{1}{2}})$
$=log_{⁡5^{−3}}(5^{−\frac{3}{2}})$
$=−\frac{1}{6}$

For $log_{\sqrt{⁡3}}\ 81:$
$=log⁡_{3^{\frac{1}{2}}}(3^4)$

$=log⁡_{3^{\frac{1}{2}}}(3^2)$
$=2$

Putting it all together:
$(−\frac{1}{6})+8−7=−\frac{1}{6}+1=\frac{5}{6}$

Therefore, the required result is $\frac{5}{6}.$