If \displaystyle \frac{1}{3}\log _{3}M+3\log _{3}N=1+\log_{0.008}5 | Filo

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Class 11 Math Algebra Logarithm and Antilogarithm

If

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

then

(a)

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

(b)

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

(c)

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

(d)

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

Correct answer: (b)

Solution:

\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 . 0 0 8}

\frac{1}{3} lo g_{3} M + 3 lo g_{3} N = 1 + \frac{lo g 5}{lo g \frac{8}{1 0 0 0}}

\frac{1}{3} lo g_{3} M + 3 lo g_{3} N = 1 + \frac{lo g 5}{lo g 8 - lo g 1 0 0 0}

\frac{1}{3} lo g_{3} M + 3 lo g_{3} N = 1 + \frac{lo g 5}{3 ( lo g 2 - lo g 1 0 )}

\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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determinants

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binomial theorem

complex numbers and quadratic equations

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Copyright FILO EDTECH PVT. LTD. 2021