If 1/3 log3 M + 3 log3 N = 1 + log 0.008 5, then

Question

If \(\frac{1}{3}\) log₃ M + 3 log₃ N = 1 + log_0.008 5, then

(a) \(M^9=\frac{9}{N}\)

(b) \(N^9=\frac{9}{M}\)

(c) \(M^3=\frac{3}{N}\)

(c) \(N^9=\frac{3}{M}\)

Answer 1

(b) \(N^9=\frac{9}{M}\)

\(\frac{1}{3}\) log₃ M + 3 log₃ N = 1 + log_0.008 5

⇒ log₃ M^1/3 + log₃N³ = 1 + log_0.0085 

⇒ log₃ M^1/3 N³ = 1 + log_0.0085 

⇒ M^1/3 N³ = 3^{(1 + log_0.0085)} 

⇒ M^1/3 N³ = 3¹ . 3^log_0.0085

⇒ \(N^9=\frac{27}{M}\big(3^{3log_{0.008}5}\big)\)

 ⇒ \(N^9=\frac{27}{M}\big(3^{log_{(0.2)^3}(5^3)}\big)\)

 ⇒ \(N^9=\frac{27}{M}\big(3^{log_{0.2}5}\big)\)                 \(\bigg[\because\,\text{log}_{a^n}x^m=\frac{m}{n}\text{log}_ax\bigg]\)

 ⇒ \(N^9=\frac{27}{M}\big(3^{log_{\frac{1}{5}}5}\big)\) = \(\frac{1}{M}\) (27)(3^-1) = \(\frac{9}{M}.\)