Question

if 3^x=5^y=15^-z then show that 1/x + 1/y +​

Answer 1

Answer:

$\mathfrak{\huge \underline{Question:}} \\\\\sf 3^x=5^y=15^-z$

$\mathfrak{\huge \underline{To Prove:}}\\\\\sf \frac{1}{x} + \frac{1}{y}+ \frac{1}{z}=0$

$\mathfrak{\huge \underline{Solution:}}$

$\sf Let~ 3^x=5^y=15^-z=K \\\\\sf Now, 3^x=K \: \: \: \longrightarrow xlog3=logK \\\\\sf \longrightarrow\frac{1}{x}=log_k 3...........[i]\\\\\sf </p><p>\huge\underline{\: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \:} \\\\\sf Now, </p><p>5^y=K \\\\\sf ylog5= logk \\\\\sf \frac{1}{y}=log_k 5...........[ii] \\\\\sf \huge\underline{\: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \:} \\\\\sf Now, 15^-z=K \\\\\sf - zlog15= logk \\\\\sf \frac{-1}{z}=log_k 15...........[iii] \\\\\sf \huge\underline{\: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \:} \\\\\sf$

From [iii],

$\sf \frac{-1}{z}=log_k 15 \\\\\sf = \frac{-1}{z}=log_k (3x5) = log_k 3 + log_k 5 \\\\\sf \frac{-1}{z}=\frac{1}{x}+\frac{1}{y} \\\\\sf \frac{1}{x}+\frac{1}{y} = \frac{-1}{z} \\\\\sf =\frac{1}{x}+\frac{1}{y} + \frac{1}{z}=0</p><p>$

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