If 1/3 log3 M + 3 log3 N = 1 + log 0.008 5, then -Maths 9th

If 1/3 log3 M + 3 log3 N = 1 + log 0.008 5, then -Maths 9th
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(b) \(N^9=rac{9}{M}\)\(rac{1}{3}\) log3 M + 3 log3 N = 1 + log0.008 5⇒ log3 M1/3 + log3N3 = 1 + log0.0085 ⇒ log3 M1/3 N3 = 1 + log0.0085 ⇒ M1/3 N3 = 3(1 + log0.0085) ⇒ M1/3 N3 = 31 . 3log0.0085⇒ \(N^9=rac{27}{M}\big(3^{3log_{0.008}5}\big)\) ⇒ \(N^9=rac{27}{M}\big(3^{log_{(0.2)^3}(5^3)}\big)\) ⇒ \(N^9=rac{27}{M}\big(3^{log_{0.2}5}\big)\)                 \(\bigg[\because\, ext{log}_{a^n}x^m=rac{m}{n} ext{log}_ax\bigg]\) ⇒ \(N^9=rac{27}{M}\big(3^{log_{rac{1}{5}}5}\big)\) = \(rac{1}{M}\) (27)(3-1) = \(rac{9}{M}.\)
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