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GuDViN [60]
3 years ago
8

Could someone please help me?

Mathematics
1 answer:
zubka84 [21]3 years ago
7 0

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, <em>not</em> 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

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lozanna [386]

Answer:

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the way you wrote this was kinda weird so I don't know what you expected as an answer

Step-by-step explanation:

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