given-27-a-64-b-216-c-72-find-2020abc-3ab-3ac-3bc-ab-ac-bc-2020abc-

Question Number 88128 by john santu last updated on 08/Apr/20

g i v e n 27^{a} = 64^{b} = 216^{c} = 72

f i n d \frac{2020 a b c}{3 a b + 3 a c + 3 b c} + \frac{a b + a c + b c}{2020 a b c}

Answered by $@ty@m123 last updated on 08/Apr/20

G i v e n 27^{a} = 64^{b} = 216^{c} = 72

27 = 72^{\frac{1}{a}} \dots (i)

64 = 72^{\frac{1}{b}} \dots (i i)

216 = 72^{\frac{1}{c}} \dots (i i i)

∴ 27 \times 64 \times 216 = 72^{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}

\Rightarrow 3^{3} \times 2^{6} \times {(2 \times 3)}^{3} = {(3^{2} \times 2^{3})}^{m} {w h e r e m = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}

\Rightarrow 3^{6} \times 2^{9} = 3^{2 m} \times 2^{3 m}

\Rightarrow m = 3 \dots (i i i)

T h e g i v e n e x p r e s s i o n

\frac{2020}{3} \times \frac{1}{m} + \frac{m}{2020}

= \frac{2020}{9} + \frac{3}{2020}

= 224 \frac{8107}{18180}

Commented by john santu last updated on 08/Apr/20

s a m e . b u t i {d o n}^{'} t u n d e r s t a n d

w h y t h e a n s w e r \frac{1}{3} ?