Answer to Question #159128 in Algebra for Jan

Question

We have seen various exponent laws for powers with the same base. CREATE and SOLVE a problem that says to 'SIMPLIFY' and that:  

a) includes multiplying powers with the same base.

b) includes dividing powers with the same base .

c) includes a power of a power.

d) includes an negative exponent.

e) includes an exponent of 0.

f) includes a rational exponent.

g) combines three of the above exponent laws. 

Answer 1

Solution

• Simplify

1)

• Question: $\qquad\qquad \begin{aligned} a^2\times a^3 \times a^\frac{2}{3} \end{aligned}$
• Answer:$\qquad\qquad \begin{aligned} &=a^{2+3+\frac{2}{3}}\\ &=a^\frac{17}{3} \end{aligned}$

2)

• Q:$\qquad\qquad \begin{aligned} \frac{(a+b)^8}{(a+b)^5} \end{aligned}$
• A:$\qquad\qquad \begin{aligned} &=(a+b)^{(8-5)}\\ &=(a+b)^3 \end{aligned}$

3)

• Q:$\qquad\qquad \begin{aligned} \sqrt{(a+b)^8} \end{aligned}$
• A:

$\qquad\qquad \begin{aligned} &=\bigg[(a+b)^8\bigg]^{\frac{1}{2}}\\ &=(a+b)^{8\times\frac{1}{2}}\\ &=(a+b)^4 \end{aligned}$

4)

• Q:$\qquad\qquad \begin{aligned} (\frac{a}{p})^{-5} \end{aligned}$
• A:

$\qquad\qquad \begin{aligned} &=\frac{a^{-5}}{p^{-5}}\\ &=\frac{1}{a^5}\times\frac{p^5}{1}\\ &=\frac{p^5}{a^5}\\ &=(\frac{p}{a})^5 \end{aligned}$

5)

• Q:$\qquad \begin{aligned} (r+p)^5 \div(r+p)^5 \end{aligned}$
• A:

$\qquad\qquad \begin{aligned} &=\frac{(r+p)^5}{(r+p)^5}\\ &=(r+p)^{5-5}\\ &=(r+p)^0\\ &=1 \end{aligned}$

f)

• Q:$\qquad\qquad \begin{aligned} \frac{a^{\frac{2}{3}}}{a^{\frac{3}{4}}}\\ \end{aligned}$
• A:

$\qquad\qquad \begin{aligned} &=a^{(\frac{2}{3}-\frac{3}{4})}\\ &=a^{-\frac{1}{12}}\\ &=\frac{1}{a^\frac{1}{12}} \end{aligned}$

g)

• Q:$\qquad \begin{aligned} \frac{a^8\times b^{\frac{3}{4}}}{(ab)^3\times c^{-6}} \end{aligned}$
• A:

$\qquad \begin{aligned} &=\frac{a^8\times b^{\frac{3}{4}}}{a^3\times b^3\times c^{-6}}\\ &=\frac{a^{(8-5)}\times b^{(\frac{3}{4}-3)}}{c^{-6}}\\ &=\frac{a^3\times b^{-\frac{9}{4}}}{c^{-6}}\\ &=\frac{a^3\times c^6}{b^{\frac{9}{4}}}\\ &=\frac{a^3\times (c^2)^3}{b^{\frac{9}{4}}}\\ &=\frac{(ac^2)^3}{b^{\frac{9}{4}}} \end{aligned}$