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Rules for Finding the Base of a Exponential Function?

RidiculousName

New member
Jun 27, 2018
28
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

Administrator
Staff member
Feb 24, 2012
13,734
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
Jan 30, 2018
365
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.