$a^{l o g_{a}^{m}} = m$ ${a, m > a n d a \neq 1}$

True

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False

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Solution

The correct option is A
True

Let's take LHS $y = a^{l o g_{a}^{m}}$

We know if $y = a^{x}, x = l o g_{a} y$

Then $l o g_{a}^{m} = l o g_{a} y$

This is possible only when $m = y$

We can say that $m = y = a^{l o g_{a}^{m}}$

So,given statement is true.

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A B C D

A B + B C + C D + D A > 2 A C

a^{2} + b^{2} - 2 (a b - a c + b c)

= (a - b) {(a - b) + 2 c} = (a - b) (a - b + 2 c)

= \frac{1}{3}

9 A D^{2} = 7 A B^{2}

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