Solve power with negative exponent $\frac{125^{6}\times 25^{-3}}{(5^{2})^{-3}\times25^{7}}$

Question

I am currently studying about exponents and powers for college calculus discipline. In the meantime I came across negative exponents, like this $25^{-3}$ and $(5^{2})^{-3}$.

I have this calculation to solve,

$$ \frac{125^{6}\times 25^{-3}}{(5^{2})^{-3}\times25^{7}} $$

But, I get very confused and I end up getting stuck in calculations with negative exponents (i.e. $b^{-a}$) and I do not know how to solve them. So I'd like to know. How can I resolve powers with negative exponents?

Answer 1

We have that

• $25=5^2$
• $125=5^3$

therefore by

• $(a^n)^m=a^{nm}$

• $a^n \times a^m = a^{n+m}$

• $\frac{a^n}{a^m} = a^{n-m}$

we have

$$\frac{125^{6}\times 25^{-3}}{(5^{2})^{-3}\times25^{7}}=\frac{5^{18}\times 5^{-6}}{5^{-6}\times 5^{14}}=\frac{5^{18-6}}{5^{14-6}}=\frac{5^{12}}{5^{8}}=5^{12-8}=5^4$$

Answer 2

The trick is to move negative exponents from top to bottum and vice versa.

Watch me do it: $$\frac{125^{6}\times 25^{-3}}{(5^{2})^{-3}\times25^{7}}$$

$$= \frac{125^{6}}{(5^{2})^{-3}\times25^{7}\times 25^{3}}$$

$$=\frac{(5^{2})^{3}\times125^{6}}{25^{7}\times 25^{3}}$$

$$\frac {5^6\times 5^{18}}{5^{14}\times 5^{6}}$$

$$= 5^4 = 625 $$

Answer 3

\begin{align} \frac{125^6\times 25^{-3}}{(5^2)^{-3}\times 25^7}= \frac{\left(\dfrac{125^6}{25^3}\right)}{\left(\dfrac{25^7}{(5^2)^3}\right)}= \dfrac{125^6}{25^3}\dfrac{(5^2)^3}{25^7}= \dfrac{(5^3)^6}{(5^2)^3}\dfrac{(5^2)^3}{(5^2)^7}= \dfrac{5^{3\times 6 + 2\times 3}}{5^{2\times 3 + 2\times 7}}= 5^{24-20}= 5^4= 625. \end{align}