Question

I need help solving and understanding this problem too!!!

Answer 1

ANSWER

D.

EXPLANATION

If

$ln(x) = y$

Then solving for x will gives us:

${e}^{ ln(x) } = {e}^{y}$

$x = {e}^{y}$

This first pair are equivalent:

If

$log_{b}(N)= P$

we

Then,

${b}^{log_{b}(N) = {b}^{ P}}$

$N= {b}^{ P}$

This pairs are also equivalent.

If

$x = \sqrt{y}$

Then we rewrite as an index to obtain;

$x = {y}^{ \frac{1}{2} }$

This pair is also equivalent.

$log_{p}(N)= b$

Then

${p}^{log_{p}(N)} = {p}^{b}$

${p}^{b} = N$

But this was not the pair given in the last option.

The correct answer is D