Question 201539
{{{f(x) = 2^x}}} means this function will take whatever you give it and raise 2 to that power. So when we give it g(x) it will raise 2 to that power: {{{f(g(x)) = 2^((g(x)))}}}<br>
So now we can see what will happen to this for the various definitions of g(x).
A. g(x)= cx
{{{f(cx) = 2^(cx)}}}
B. g(x)= c/x
{{{f(c/x) = 2^(c/x) = 2^(c*(1/x))}}}
C. g(x)= x/c
{{{f(x/c) = 2^(x/c) = 2^(x*(1/c))}}}
D. g(x)= x-c
{{{f(x-c) = 2^((x-c))}}}
E. g(x)= log x
{{{f(log((x))) = 2^(log((x)))}}}<br>
Now we just need to figure out which exponent is largest
When c>1 and x>1 ...
A. cx > x and cx > c
B. 1/x is between 0 and 1 so c*(1/x) < c.
C. 1/c is between 0 and 1 so x*(1/c) < x.
D. x-c < x
E. log(x) < x. This is the trickiest one to explain. If you look at graphs of y=x and y=log(x) superimposed on each other you will see that for all x-values the graph of graph of y=log(x) is below the graph of y=x. So log(x) < x for all x. (Unfortunately I can't get Algebra.com software to graph log(x). If I could I'd show you these graphs here.)<br>
In summary, when c>1 and x>1 only c*x is guaranteed to be larger than both c and x. So c*x is the largest exponent which means {{{2^(cx)}}} is the largest which means the answer is A.