Question

1. Use the Laws of Exponents to rewrite and simplify each expression.

(a) \(\frac{{ - {2^6}}}{{{4^3}}}\)

(b) \(\frac{{{{\left( { - {\bf{3}}} \right)}^6}}}{{{{\bf{9}}^{\bf{6}}}}}\)

(c) \(\frac{{\bf{1}}}{{\sqrt({\bf{4}}){{{x^{\bf{5}}}}}}}\)

(d) \(\frac{{{x^{\bf{3}}} \cdot {x^n}}}{{{x^{n + 1}}}}\)

(e) \({b^{\bf{3}}}{\left( {3{b^{ - {\bf{1}}}}} \right)^{ - {\bf{2}}}}\)

(f) \(\frac{{{\bf{2}}{x^{\bf{2}}}y}}{{{{\left( {3{x^{ - {\bf{2}}}}y} \right)}^{\bf{2}}}}}\)

Short answer

(a) \(\frac{{ - {2^6}}}{{{4^3}}} = - 1\)

(b) \(\frac{{{{\left( { - 3} \right)}^6}}}{{{9^6}}} = \frac{1}{{{3^6}}}\)

(c) \(\frac{1}{{\sqrt(4){{{x^5}}}}} = \frac{1}{{x\sqrt(4){x}}}\)

(d) \(\frac{{{x^3} \cdot {x^n}}}{{{x^{n + 1}}}} = {x^2}\)

(e) \({b^3}{\left( {3{b^{ - 1}}} \right)^{ - 2}} = \frac{{{b^5}}}{9}\)

(f) \(\frac{{2{x^2}y}}{{{{\left( {3{x^{ - 2}}y} \right)}^2}}} = \frac{{2{x^6}}}{{9y}}\)

Step-by-step solution

Simplify the expression (a)

The given expression is \(\frac{{ - {2^6}}}{{{4^3}}}\).

Use the law of exponents\({\left( {{b^x}} \right)^y} = {b^{xy}}\), and\(\frac{{{b^x}}}{{{b^y}}} = {b^{x - y}}\)to simplify the above expression as shown below:

\(\begin{aligned}\frac{{ - {2^6}}}{{{4^3}}} &= \frac{{ - {2^6}}}{{{{\left( {{2^2}} \right)}^3}}}\\ &= \frac{{ - {2^6}}}{{{2^{2 \cdot 3}}}}\\ &= \frac{{ - {2^6}}}{{{2^6}}}\end{aligned}\)

Simplify further,

\(\begin{aligned}\frac{{ - {2^6}}}{{{4^3}}} &= - {\left( 2 \right)^{6 - 6}}\\ &= - {2^0}\\ &= - 1\end{aligned}\)

Thus, \(\frac{{ - {2^6}}}{{{4^3}}} = - 1\).

Simplify the expression (b)

The given expression is \(\frac{{{{\left( { - 3} \right)}^6}}}{{{9^6}}}\).

Use the law of exponents\({\left( {{b^x}} \right)^y} = {b^{xy}}\), and\(\frac{{{a^x}}}{{{b^x}}} = {\left( {\frac{a}{b}} \right)^x}\)to simplify the above expression as shown below:

\(\begin{aligned}\frac{{{{\left( { - 3} \right)}^6}}}{{{9^6}}} &= \frac{{{{\left( { - 3} \right)}^6}}}{{{{\left( 9 \right)}^6}}}\\ &= {\left( { - \frac{3}{9}} \right)^6}\\ &= {\left( { - \frac{1}{3}} \right)^6}\end{aligned}\)

Simplify further,

\(\frac{{{{\left( { - 3} \right)}^6}}}{{{9^6}}} = \frac{1}{{{3^6}}}\)

Thus, \(\frac{{{{\left( { - 3} \right)}^6}}}{{{9^6}}} = \frac{1}{{{3^6}}}\).

Simplify the expression (c)

The given expression is \(\frac{1}{{\sqrt(4){{{x^5}}}}}\).

Use the law of exponent\({b^{x + y}} = {b^x} \cdot {b^y}\)to simplify the above expression as shown below:

\(\begin{aligned}\frac{1}{{\sqrt(4){{{x^5}}}}} &= \frac{1}{{\sqrt(4){{{x^{4 + 1}}}}}}\\ &= \frac{1}{{\sqrt(4){{{x^4} \cdot x}}}}\\ &= \frac{1}{{{{\left( {{x^4}} \right)}^{1/4}} \cdot {{\left( x \right)}^{1/4}}}}\end{aligned}\)

Simplify further,

\(\begin{aligned}\frac{1}{{\sqrt(4){{{x^5}}}}} &= \frac{1}{{{{\left( {{x^4}} \right)}^{1/4}} \cdot {{\left( x \right)}^{1/4}}}}\\ &= \frac{1}{{x\sqrt(4){x}}}\end{aligned}\)

Thus, \(\frac{1}{{\sqrt(4){{{x^5}}}}} = \frac{1}{{x\sqrt(4){x}}}\).

Simplify the expression (d)

The given expression is \(\frac{{{x^3} \cdot {x^n}}}{{{x^{n + 1}}}}\).

Use the law of exponents\({b^x} \cdot {b^y} = {b^{x + y}}\), and\(\frac{{{b^x}}}{{{b^y}}} = {b^{x - y}}\)to simplify the above expression as shown below:

\(\frac{{{x^3} \cdot {x^n}}}{{{x^{n + 1}}}} = \frac{{{x^{3 + }}^n}}{{{x^{n + 1}}}}\)

Simplify further,

\(\begin{aligned}\frac{{{x^3} \cdot {x^n}}}{{{x^{n + 1}}}} &= \frac{{{x^{3 + }}^n}}{{{x^{n + 1}}}}\\ &= {x^{3 + }}^{n - \left( {n + 1} \right)}\\ &= {x^{3 + n - n - 1}}\\ &= {x^2}\end{aligned}\)

Thus, \(\frac{{{x^3} \cdot {x^n}}}{{{x^{n + 1}}}} = {x^2}\).

Simplify the expression (e)

The given expression is \({b^3}{\left( {3{b^{ - 1}}} \right)^{ - 2}}\).

Use the law of exponents\({\left( {{b^x}} \right)^y} = {b^{xy}}\), and\({b^x} \cdot {b^y} = {b^{x + y}}\)to simplify the above expression as shown below:

\(\begin{aligned}{b^3}{\left( {3{b^{ - 1}}} \right)^{ - 2}} &= {b^3}{3^{ - 2}}{\left( {{b^{ - 1}}} \right)^{ - 2}}\\ &= \frac{{{b^3} \cdot {b^2}}}{{{3^2}}}\end{aligned}\)

Simplify further,

\(\begin{aligned}{b^3}{\left( {3{b^{ - 1}}} \right)^{ - 2}} &= \frac{{{b^3} \cdot {b^2}}}{{{3^2}}}\\ &= \frac{{{b^5}}}{9}\end{aligned}\)

Thus,\({b^3}{\left( {3{b^{ - 1}}} \right)^{ - 2}} = \frac{{{b^5}}}{9}\).

Simplify the expression (f)

The given expression is \(\frac{{2{x^2}y}}{{{{\left( {3{x^{ - 2}}y} \right)}^2}}}\).

Use the law of exponents\({\left( {{b^x}} \right)^y} = {b^{xy}}\), and\({b^x} \cdot {b^y} = {b^{x + y}}\)to simplify the above expression as shown below:

\(\begin{aligned}\frac{{2{x^2}y}}{{{{\left( {3{x^{ - 2}}y} \right)}^2}}} &= \frac{{2{x^2}y}}{{{3^2}{{\left( {{x^{ - 2}}} \right)}^2}{y^2}}}\\ &= \frac{{2{x^2}y}}{{9{x^{ - 4}}{y^2}}}\end{aligned}\)

Simplify further,

\(\begin{aligned}\frac{{2{x^2}y}}{{{{\left( {3{x^{ - 2}}y} \right)}^2}}} &= \frac{{2{x^2}y}}{{9{x^{ - 4}}{y^2}}}\\ &= \frac{2}{9}{x^{2 - \left( { - 4} \right)}}{y^{1 - 2}}\\ &= \frac{2}{9}{x^6}{y^{ - 1}}\\ &= \frac{{2{x^6}}}{{9y}}\end{aligned}\)

Thus,\(\frac{{2{x^2}y}}{{{{\left( {3{x^{ - 2}}y} \right)}^2}}} = \frac{{2{x^6}}}{{9y}}\).