I'm having trouble simplifying this:
(2x^-2)^2(3x^4)^-3
the answer is 4/27x^16
but I have no idea how they got this??
Also, solving for 3a^(4/5)=48 could you please guide me through the steps?
thank you so much!
I'm having trouble simplifying this:
(2x^-2)^2(3x^4)^-3
the answer is 4/27x^16
but I have no idea how they got this??
Also, solving for 3a^(4/5)=48 could you please guide me through the steps?
thank you so much!
sorry, forgot to show my steps so far:
for the first qu:
I have it up to: 1/(4x^4)(-27x^12)
I don't understand how they got a 4 over 27x^16? I would think it would be a 1 over 27x^16.
For the second question:
I have 5sqrt(3a^4) = 48
and then I'm sorta stuck... could it be: 5*4=20, so (a^20)/3 = 16?
use the exponent laws $\displaystyle \left(x^{a}\right)^{b}=x^{ab}$ and $\displaystyle (xy)^{a}=x^{a}y^{b}$:
$\displaystyle \left(2x^{-2}\right)^{2}\left(3x^4\right)^{-3}=2^2x^{-4}3^{-3}x^{-12}$
then remember that $\displaystyle x^{a}x^{b}=x^{a+b}$:
$\displaystyle =2^2\cdot 3^{-3}x^{-16}$
now recall that $\displaystyle x^{-a}=\left(\frac{1}{x^{a}}\right)$. Also note that $\displaystyle 3^{-3}$ is a positive number:
$\displaystyle =2^2\left(\frac{1}{3^3}\right)\left(\frac{1}{x^{16 }}\right)=\frac{4}{27x^{16}}$