Question

For some positive real number $x$ , if $log_{\sqrt 3}(x)+\frac{log_x(25)}{log_x(0.008)}=\frac{16}{3}$, then the value of $log_3(3x^2)$ is

• A. 4
• B. 6
• C. 7
• D. 9

Answer 1

Answer: (C)

7

Answer 2

Given : $\log_{\sqrt3}(x)+\frac{\log_x(25)}{\log_x(0.008)}=\frac{16}{3}$, which can be stated as :

⇒ 2 log3x + log0.0825 = $\frac{16}{3}$

⇒ 2 log3x + $\log_{\frac{8}{1000}}$25 = $\frac{16}{3}$

⇒ 2 log3x + $\log_{5^{-3}}(5)^2=\frac{16}{3}$

⇒ 2 log3x - $\frac{2}{3}=\frac{16}{3}$

⇒ 2 log3x = $\frac{16}{3}+\frac{2}{3}$
⇒ 2 log3x = 6
⇒ log3x2 = 6
⇒ x2 = 36
Therefore, log3 (3.x2)
= log3 (3.36)
= log3 37
= 7

So, the correct option is (C): 7.