1 x + 1 y + 1 z \frac{1}{x}+\frac{1}{y}+\frac{1}{z}

Algebra Level 2

a a , b b and c c are positive real numbers that satisfy a x = b y = c z = 125 , log ⁡ 5 a b c = 3. a^x=b^y=c^z=125, \log_{5} abc=3. What is the value of 1 x + 1 y + 1 z ? \frac{1}{x}+\frac{1}{y}+\frac{1}{z}?

1 3 \frac{1}{3} 1 1 1 15 \frac{1}{15} 1 5 \frac{1}{5}

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Mahindra Jain by Mahindra Jain

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1 solution

We have, a = 12 5 1 x , b = 12 5 1 y , c = 12 5 1 z a = 125^{\frac{1}{x}} , b = 125^{\frac{1}{y}} , c = 125^{\frac{1}{z}}

a b c = 12 5 1 x × 12 5 1 y × 12 5 1 z = 12 5 1 x + 1 y + 1 z abc = 125^{\frac{1}{x}} \times 125^{\frac{1}{y}} \times 125^{\frac{1}{z}} = 125^{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}

Taking log ⁡ 5 \log_5 both sides,

log ⁡ 5 a b c = log ⁡ 5 12 5 1 x + 1 y + 1 z \log_5{abc} = \log_5{125^{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}}

3 = ( 1 x + 1 y + 1 z ) log ⁡ 5 125 3 = (\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) \log_5{125}

3 = ( 1 x + 1 y + 1 z ) × 3 3 = (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})\times 3

1 x + 1 y + 1 z = 1 \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \boxed{1}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

you rite

Chanchal Ahamed - 7 years, 3 months ago

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