Rules for Finding the Base of a Exponential Function?

[RidiculousName]

I was wondering if anyone could point me to a set of rules for finding the base of an exponential function? I can figure out that the base of $$f(x)=7^x$$ is 7 and the base of $$f(x)=3^{2x}$$ is 9 but even though I know $$f(x)=8^{\frac{4}{3}x}$$ has a base of 16, I don't know how that answer was reached.

 

[MarkFL]

You could write:

$$f(x)=8^{\Large\frac{4}{3}x}=\left(8^{\Large\frac{4}{3}}\right)^x=\left(\left(8^{\Large\frac{1}{3}}\right)^4\right)^x=\left(2^4\right)^x=16^x$$

 

[HOI]

I was wondering if anyone could point me to a set of rules for finding the base of an exponential function? I can figure out that the base of $$f(x)=7^x$$ is 7 and the base of $$f(x)=3^{2x}$$ is 9 but even though I know $$f(x)=8^{\frac{4}{3}x}$$ has a base of 16, I don't know how that answer was reached.

Actually, the base of [math]3^{2x}[/math] is 3! Of course that is the equal to [math](3^2)^x= 9^x[/math] which base 9. [math]f(x)= 8^{\frac{4}{3}ax}[/math] has base 8. Using the "laws of exponents", [math]8^{4/3}= (8^{1/3})^4[/math] and, since [math]2^3= 8[/math], [math]8^{1/3}= 2[/math] so [math]8^{4/3}= 2^4= 16[/math] so that [math]8^{\frac{4}{3}x}= 16^x[/math].

But, again, I would say that the bases have changed. The base in [math]8^{\frac{4}{3}x}[/math] is 8 and the base in [math]16^x[/math] is 16.