Question

What is the simplified base for the function f(x) = 2(27 cube rooted to the power of 2x

Answer 1

Remember: $(a^b)^c=a^{bc}$ and $\sqrt[n]{a^b}=a^\frac{b}{n}$

Not sure if you mean

AAA. $f(x)=2(\sqrt[3]{27})^{2x}$

or

BBB. $2(\sqrt[3]{27^{2x}})$

Or

CCC. $(2\sqrt[3]{27})^{2x}$

If AAA, go to AAAAAA

If BBB, go to BBBBB

If CCC, go to CCCCC

AAAAAAAAA

$f(x)=2(\sqrt[3]{27})^{2x}$

Simplify inside parenthaees first

$f(x)=2(\sqrt[3]{3^3})^{2x}$

$f(x)=2(3)^{2x}$

$f(x)=2((3)^2)^x$

$f(x)=2(9)^x$

The base is 9

BBBBBBBB

$f(x)=2(\sqrt[3]{27^{2x})$

$f(x)=2(\sqrt[3]{(3^3)^{2x})$

$f(x)=2(\sqrt[3}{3^{6x}})$

$f(x)=2(3^\frac{6x}{3})$

$f(x)=2(3^{2x})$

$f(x)=2(3^2)^x$

$f(x)=2(9)^x$

The base is 9

CCCCCCC

$f(x)=(2\sqrt[3]{27})^{2x}$

$f(x)=(2\sqrt[3]{3^3})^{2x}$

$f(x)=(2*3)^{2x}$

$f(x)=6^{2x}$

$f(x)=(6^2)^x$

$f(x)=36^x$

Base is 36

If it’s $f(x)=2(\sqrt[3]{27})^{2x}$ or $f(x)=2(\sqrt[3]{27^{2x}})$, the base is 9

If it’s $f(x)=(2\sqrt[3]{27})^{2x}$, the base is 36