Solve $3^{1/4} \cdot 9^{-5/8}$

Question

I don't understand how to solve $3^{1/4} \cdot 9^{-5/8}$. Help please?

I have tried many different things, but they're not working. Once I plug the problem into a math equation solver, the answer $1/3$ appears, but I don't understand how they got that.

Answer 1

Using $\displaystyle(a^m)^n=a^{mn}$ where $m,n$ are real

$\displaystyle9^{-\frac58}=(3^2)^{-\frac58}=(3)^{-\frac54}$

and using $\displaystyle a^m\cdot a^n=a^{m+n}$

$$\displaystyle3^{\frac14}\cdot(9)^{-\frac58}=3^{\left(\frac14-\frac54\right)}=\cdots$$

See Exponent Combination Laws

Answer 2

The key to this is that $9=3^2$. That gives us:

$$ 3^{1/4} \cdot (3^2)^{-5/8}$$

We know that $(a^m)^n = a^{mn}$, so then we have $$3^{1/4} \cdot 3^{2 \cdot -5/8} = 3^{1/4} \cdot 3^{-10/8}$$

It helps to simplify fractions: $$= 3^{1/4} \cdot 3^{-5/4}$$

We know that $a^m \cdot a^n = a^{m+n}$, so:

$$3^{1/4} \cdot 3^{-5/4} = 3^{-4/4} = 3^{-1} = \frac13$$