Question

Could someone please help me?

Answer 1

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

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 \\eq 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`