Question

Could someone please help me?

Answer 1

Answer:

A. $Log_p N = b \ and \ b^p = N$

Step-by-step explanation:

A. The given options are;

$Log_p N = b \ and \ b^p = N$

For $Log_p N = b$ we have;

$p^b = N$

However;

$b^p = N$

Where, p ≠ b

$p^b \neq b^p$

Therefore;

N ≠ N for the given pair and the pair are therefore, not equivalent

B. x = √y and $x = y^{1/2}$

We note that

x = √y = $y^{1/2}$

We have;

x = $y^{1/2}$, by transitive property

Therefore, the pair, x = √y and $x = y^{1/2}$ are equivalent

C. $Log_b N = p \ and \ b^p = N$

From $Log_b N = p$, we have;

$b^p = N$

Therefore, $Log_b N = p \ and \ b^p = N$ are equivalent pairs as $b^p = N$ can be obtained from $Log_b N = p$

D. ㏑x = y and x = $e^y$

$ln \ x = ln_e \ x = y$, therefore, by rules of logarithm, we have;

x = $e^y$

Therefore, ㏑x = y and x = $e^y$ are equivalent pairs, as x = $e^y$ can be obtained from ln x.

Therefore;

The option which are not equivalent pair is option is $Log_p N = b \ and \ b^p = N$