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`7^(log_(3)5)+3^(log_(5)7)-5^(log_(3)7)-7^(log_(5)3)` |
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Answer» 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, `3^(log_(5)7) = 7^(log_(5)3)`->(2) So, using (1) and (2), we can write our expression as, `5^(log_(3)7)+7^(log_(5)3)- 5^(log_(3)7)-7^(log_(5)3) = 0` |
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