If 31log3M+3log3N=1+log0.0085 then

31log3M+3log3N=1+log0.0085 31log3M+3log3N=1+log0.008log5 31log3M+3log3N=1+log10008log 5 31log3M+3log3N=1+log8−log1000log 5 31log3M+3log3N=1+3(log2−log10)log 5 31log3M+3log3N=1+3(log51log 5 log3M31N3=1−31 logM31N3=32log3M31 N3 =332MN9=9N9=M9

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If {1}{3} _{3}M+3 _{3}N=1+_{0.008}5 then

Question

If $\frac{1}{3} lo g_{3} M + 3 lo g_{3} N = 1 + lo g_{0.008} 5$ then

M^{9} = \frac{9}{N}

N^{9} = \frac{9}{M}

M^{3} = \frac{3}{N}

N^{9} = \frac{3}{M}

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\frac{1}{3} lo g_{3} M + 3 lo g_{3} N = 1 + lo g_{0.008} 5

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Text solutionVerified

\frac{1}{3} lo g_{3} M + 3 lo g_{3} N = 1 + lo g_{0.008} 5

\frac{1}{3} lo g_{3} M + 3 lo g_{3} N = 1 + \frac{lo g 5}{lo g 0.008}

\frac{1}{3} lo g_{3} M + 3 lo g_{3} N = 1 + \frac{lo g 5}{lo g \frac{8}{1000}}

\frac{1}{3} lo g_{3} M + 3 lo g_{3} N = 1 + \frac{lo g 5}{lo g 8 - lo g 1000}

\frac{1}{3} lo g_{3} M + 3 lo g_{3} N = 1 + \frac{lo g 5}{3 ( lo g 2 - lo g 10 )}

\frac{1}{3} lo g_{3} M + 3 lo g_{3} N = 1 + \frac{lo g 5}{3 ( lo g \frac{1}{5}}

lo g_{3} M^{\frac{1}{3}} N^{3} = 1 - \frac{1}{3}

lo g M^{\frac{1}{3}} N^{3} = \frac{2}{3} lo g 3

M^{\frac{1}{3}} N^{3} = 3^{\frac{2}{3}}

M N^{9} = 9

N^{9} = \frac{9}{M}

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Question Text	If $\frac{1}{3} lo g_{3} M + 3 lo g_{3} N = 1 + lo g_{0.008} 5$ then
Answer Type	Text solution:1
Upvotes	150

If 31​log3​M+3log3​N=1+log0.008​5 then

Text solutionVerified

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If $\frac{1}{3} lo g_{3} M + 3 lo g_{3} N = 1 + lo g_{0.008} 5$ then