Answer:
A. [tex]Log_p N = b \ and \ b^p = N[/tex]
Step-by-step explanation:
A. The given options are;
[tex]Log_p N = b \ and \ b^p = N[/tex]
For [tex]Log_p N = b[/tex] we have;
[tex]p^b = N[/tex]
However;
[tex]b^p = N[/tex]
Where, p ≠ b
[tex]p^b \neq b^p[/tex]
Therefore;
N ≠ N for the given pair and the pair are therefore, not equivalent
B. x = √y and [tex]x = y^{1/2}[/tex]
We note that
x = √y = [tex]y^{1/2}[/tex]
We have;
x = [tex]y^{1/2}[/tex], by transitive property
Therefore, the pair, x = √y and [tex]x = y^{1/2}[/tex] are equivalent
C. [tex]Log_b N = p \ and \ b^p = N[/tex]
From [tex]Log_b N = p[/tex], we have;
[tex]b^p = N[/tex]
Therefore, [tex]Log_b N = p \ and \ b^p = N[/tex] are equivalent pairs as [tex]b^p = N[/tex] can be obtained from [tex]Log_b N = p[/tex]
D. ㏑x = y and x = [tex]e^y[/tex]
[tex]ln \ x = ln_e \ x = y[/tex], therefore, by rules of logarithm, we have;
x = [tex]e^y[/tex]
Therefore, ㏑x = y and x = [tex]e^y[/tex] are equivalent pairs, as x = [tex]e^y[/tex] can be obtained from ln x.
Therefore;
The option which are not equivalent pair is option is [tex]Log_p N = b \ and \ b^p = N[/tex]
