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7^(log(3)5)+3^(log(5)7)-5^(log(3)7) ...

`7^(log_(3)5)+3^(log_(5)7)-5^(log_(3)7) -7^(log_(5)3)`

Text Solution

Verified by Experts

We know, `log_ab = log_xb/log_xa`
So, `7^(log_(3)5) =7^(log_(7)5/(log_(7)3) `
`= (7^(log_(7)5))^(1/log_(7)3)` (As `a^(log_(a)b) = b`)
`=5^((1/log_(7)3)` (As `a^(log_(a)b) = b`)
`= 5^(log_(3)7)` (As `a^(1/log_(b)c) = a^(log_(c)b)`)
So,`7^(log_(3)5) = 5^(log_(3)7)`->(1)
Similarly, we can show that,
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