c > 1.If f(x) = 2

^{x}, which of these 5 functions yields the greatest value for f(g(x)), for all

x > 1?

B. g(x) = c/x

C. g(x) = x/c

D. g(x) = x – c

E. g(x) = log

_{c}x

Listed below are 5 functions, each denoted g(x) and each involving a real number constant

c > 1.If f(x) = 2^{x}, which of these 5 functions yields the greatest value for f(g(x)), for all

x > 1?

c > 1.If f(x) = 2

x > 1?

A. g(x) = cx

B. g(x) = c/x

C. g(x) = x/c

D. g(x) = x – c

E. g(x) = log_{c}x

B. g(x) = c/x

C. g(x) = x/c

D. g(x) = x – c

E. g(x) = log

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John M. | John - Algebra TutorJohn - Algebra Tutor

f(x)=2^{x}

Then f(g(x))= 2^{g(x)}

we are supposed evalute the 5 expressions to determine which would yield the highest value of f(x). Because the g(x) functions share a common base, we can evalute which one would have the highest value, and that would yeild the highest value for f(x).^{
}

For any value of c, answer A. is clearly larger than a fraction (B. and C.), and larger than D. The only sticky point is answer E, is cx larger than log_{c}x?

-------

In general the expression: log_{a}b=n can be written as a^{n}=b

For example: let a = 2 and b=8. We know that n=3 because 8 is 2^{3}. Similarly, log_{2}8=3.

------

returning to the problem, we know that the value of log_{c}x must be smaller than x for any c, therefore it is less than "cx" as well.

So the answer is: A. g(x)=cx, yeilds the highest value of f(g(x)) for all x>1.

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