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3 years ago

Assume a is a positive constant. Imagine solving for x (but do not actually do so). Will your answer involve logarithms?

Mathematics

1 answer:

finlep [7]3 years ago

5 0

9514 1404 393

Answer:

(b) No, 10^a is involved

Step-by-step explanation:

The relationship of interest is ...

$a=\log{(x)}\ \leftrightarrow\ 10^a=x$

Determining x involves the exponential 10^a.

_____

When faced with problems of this sort, I like to remind myself that a logarithm is an exponent.

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Step-by-step explanation:

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3 years ago

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4 years ago