Math, asked by yashanakhare, 1 year ago

if 2^x = 5^y = 40^y, then prove that 1/z = 3/x + 1/y

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Answered by Anonymous

$HEYA \\ \\ \\ GIVEN \: \: QUESTION \: \: Is \: \\ \\ 2 {}^{x} = 5 {}^{y} = 40 {}^{z} \\ \\ let \: \: \: 2 {}^{x} = 5 {}^{y} = 40 {}^{z} = k \\ \\ 2 {}^{x} = k \: \: \: \: \: 5 {}^{y} = k \: \: \: \: 40 {}^{z} = k \\ \\ 2 = k {}^{ \frac{1}{x} } \: \: \: .... \: Equation \: \: i \\ \\ 5 = k {}^{ \frac{1}{y} } \: \: \: ... \: \: Equation \: \: \: ii \\ \\ 40 = k {}^{ \frac{1}{z} } \: \: \: .... \: \: Equation \: \: \: iii \\ \\ \\ 40 = k {}^{ \frac{1}{z} } \\ \\ 5 \times 2 {}^{3} = k {}^{ \frac{1}{z} } \\ \\ k {}^{ \frac{1}{y} } \times k {}^{ \frac{1}{3x} } = k {}^{ \frac{1}{z} } \: \: \: \: by \: using \: Equation \: i \: and \: ii \\ \\ k {}^{ (\frac{1}{y} + \frac{1}{3x} ) } = k {}^{ \frac{1}{z} } \\ \\ now \: compare \: powers \: of \: k \: we \: have \\ \\ \frac{1}{y} + \frac{1}{3x} = \frac{1}{z} \: \: \: \: hence \: \: proved$

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if 2^x = 5^y = 40^y, then prove that 1/z = 3/x + 1/y​

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if 2^x = 5^y = 40^y, then prove that 1/z = 3/x + 1/y