Question

if a^x=(a/k)^y=k^m then 1/x_1/y=?

Answer 1

a^x = (a/k)^y = k^m = P ( Let)
a^x = P
take log both sides,
xloga = logP
x = logP/loga ------------(1)

similarly ,
(a/k)^y = P
take log both sides ,
ylog(a/k) = logP
y = logP/log(a/k) ----------(2)

k^m = P
mlogk = logP
m = logP/logk--------(3)

now, 1/x - 1/y = 1/{logP/loga} - 1/{logP/log(a/k}
= loga/logP - log(a/k)/logP
={loga - log(a/k)}/logP
={loga - loga + logk}/logP
= logk/logP
= 1/{logP/logK}
from equation (3)
= 1/m
hence , 1/x - 1/y = 1/m

Answer 2

✍✍HERE IS YOUR ANSWER..⬇⬇
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$\underline{SOLUTION}$ ➡

$\underline{Given \: : }\: \: \: a {}^{x} = (\frac{a}{k} ) {}^{y} = k {}^{m} \\ \\ \underline{ To \: \: find \: :} \: \: \frac{1}{x} - \frac{1}{y} =$

$Let ,\\ \\ a {}^{x} = \frac{a}{k} {}^{y} = {k}^{m} = z$

$So ,\\ \\ a {}^{x} = z \\ \\ = > (a {}^{x} ) {}^{ \frac{1}{x} } = z {}^{ \frac{1}{x} } \\ \\ = > a = z {}^{ \frac{1}{x} } \: \: - - - - > (1) \\ \\ Similarly \\ \\ \frac{a}{k} {}^{y} = z \\ \\ = > \frac{a}{k} = z {}^{ \frac{1}{y} } \: \: - - - - > (2) \\ \\ And ,\\ \\ k {}^{m} = z \\ \\ = > k = z {}^{ \frac{1}{m} } \: \: - - - - > (3)$

$Now, \\ \\ \frac{z {}^{ \frac{1}{x} } }{z {}^{ \frac{1}{y} } } = \frac{a}{ \frac{a}{k} } \: \: \: \: \: \: [ from \: (1)\: \: and \: \: (2)] \\ \\ = > z {}^{ \frac{1}{x} - \frac{1}{y} } = \frac{a}{a} \times k \\ \\ = > z {}^{ \frac{1}{x} - \frac{1}{y} } =k \\ \\ = > z {}^{ \frac{1}{x} - \frac{1}{y} } = z {}^{ \frac{1}{m} } \: \: \: \: \: \: \: \: [from \: \: (3)]\\ \\ = > \frac{1}{x} - \frac{1}{y} = \frac{1}{m}$

$\underline{ANSWER}$ ➡

$\boxed{ \frac{1}{x} - \frac{1}{y} = \frac{1}{m}}$

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