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प्रश्न
If ax = by = cz and abc = 1, show that
`(1)/x + (1)/y + (1)/z` = 0.
योग
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उत्तर
ax = by = cz
So, ax = by ⇒ a =`"b"^(y/x) .....("Using" "a"^(1/"n") = root("n")("a"))`
by = cz ⇒ c = `"b"^(y/z) .....("Using" "a"^(1/"n") = root("n")("a"))`
and abc = 1
⇒ `"b"^(y/x) · "b"·"b"^(y/z)` = 1
⇒ `"b"^(y/x) · "b"·"b"^(y/z)` = 1
⇒ `"b"^(y/x + 1 + y/z)` = b° ......(Using a° = 1)
⇒ `y/x + 1 + y/z` = 0
Divide throughout by y.
⇒ `(1)/x + (1)/y + (1)/z` = 0
Hence proved.
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Solving Exponential Equations
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