Question
If $a b + b \log a - 1 = 0,$ then prove that $b^a.a^b = 10$
✓
Answer
Given $a \log b + b \log a - 1 = 0$
$\Rightarrow \log(b)^a + \log(a)^b- \log 10 = 0$
$\Rightarrow \log(b^a . a^b) = \log 10$
$\therefore b^a . a^b = 10.$
$\Rightarrow \log(b)^a + \log(a)^b- \log 10 = 0$
$\Rightarrow \log(b^a . a^b) = \log 10$
$\therefore b^a . a^b = 10.$
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