# Rules for Finding the Base of a Exponential Function?

#### RidiculousName

##### New member
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 $$\displaystyle f(x)=7^x$$ is 7 and the base of $$\displaystyle f(x)=3^{2x}$$ is 9 but even though I know $$\displaystyle f(x)=8^{\frac{4}{3}x}$$ has a base of 16, I don't know how that answer was reached.

#### MarkFL

Staff member
You could write:

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

#### Country Boy

##### Well-known member
MHB Math Helper
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 $$\displaystyle f(x)=7^x$$ is 7 and the base of $$\displaystyle f(x)=3^{2x}$$ is 9 but even though I know $$\displaystyle f(x)=8^{\frac{4}{3}x}$$ has a base of 16, I don't know how that answer was reached.
Actually, the base of $$\displaystyle 3^{2x}$$ is 3! Of course that is the equal to $$\displaystyle (3^2)^x= 9^x$$ which base 9. $$\displaystyle f(x)= 8^{\frac{4}{3}ax}$$ has base 8. Using the "laws of exponents", $$\displaystyle 8^{4/3}= (8^{1/3})^4$$ and, since $$\displaystyle 2^3= 8$$, $$\displaystyle 8^{1/3}= 2$$ so $$\displaystyle 8^{4/3}= 2^4= 16$$ so that $$\displaystyle 8^{\frac{4}{3}x}= 16^x$$.

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