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Algebra Level 2

2 log ⁡ 4 200 0 6 + 3 log ⁡ 5 200 0 6 \large \dfrac{2}{\log_4 2000^6 }+ \dfrac{3}{\log_5 2000^6}

The sum above can be written as m n \dfrac{m}{n} , where m m and n n are prime positive integers. Find m + n m+n .

Note : Don't use a calculator.


The answer is 7.

Ikhwan Norazam by Ikhwan Norazam , 16, Malaysia

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1 solution

Chew-Seong Cheong
Nov 17, 2017

S = 2 log ⁡ 4 200 0 6 + 3 log ⁡ 5 200 0 6 = 2 log ⁡ 200 0 6 log ⁡ 4 + 3 log ⁡ 200 0 6 log ⁡ 5 = 2 log ⁡ 4 6 log ⁡ 2000 + 3 log ⁡ 5 6 log ⁡ 2000 = log ⁡ ( 4 2 × 5 3 ) 6 log ⁡ 2000 = log ⁡ 2000 6 log ⁡ 2000 = 1 6 \begin{aligned} S & = \frac 2{\log_4 2000^6} + \frac 3{\log_5 2000^6} \\ & = \frac 2{\frac {\log2000^6}{\log 4}} + \frac 3{\frac {\log2000^6}{\log 5}} \\ & = \frac {2\log 4}{6\log2000} + \frac {3\log 5}{6\log2000} \\ & = \frac {\log (4^2\times 5^3)}{6\log2000} \\ & = \frac {\log 2000}{6\log2000} \\ & = \frac 16 \end{aligned}

⟹ m + n = 1 + 6 = 7 \implies m+n = 1+6=\boxed{7}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

thanks for the solution it,s was very helpful

Ikhwan Norazam - 3 years, 6 months ago

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You are welcome.

Chew-Seong Cheong - 3 years, 6 months ago

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