##### Rules of Exponents

Zero-Exponent Rule:   $\displaystyle {{a}^{0}}=1$  this means that each number raised to the zero power is 1.

Worked example 1:

50 = 1

(-3)0 = 1

$\displaystyle {{(5{{x}^{6}}{{y}^{2}})}^{0}}=1$

$\displaystyle {{\left( \frac{2}{x} \right)}^{0}}=1$

Power of a Power$\displaystyle {{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$

So, when you raise a power to a power you have to multiply the exponents.

Worked example 2:

$\displaystyle {{\left( {{2}^{3}} \right)}^{4}}={{2}^{12}}$

$\displaystyle {{\left( {{x}^{2}} \right)}^{5}}={{x}^{10}}$

Negative Exponent Rule: $\displaystyle {{a}^{-n}}=\frac{1}{{{a}^{n}}}$ this means that negative exponents in the numerator get moved to the denominator and become positive exponents.

Worked example 3:

$\displaystyle {{\left( 8 \right)}^{-1}}=\frac{1}{8}$

$\displaystyle {{\left( -2 \right)}^{-3}}=\frac{1}{{{\left( -2 \right)}^{3}}}=-\frac{1}{8}$.

Product Rule: $\displaystyle {{a}^{n}}\cdot {{a}^{m}}={{a}^{n+m}}$ so,  to multiply two exponents with the same base, you keep the base and add the powers.

Worked example 4:

$\displaystyle {{2}^{3}}\cdot {{2}^{5}}={{2}^{3+5}}={{2}^{8}}$

$\displaystyle {{x}^{3}}\cdot x={{x}^{3+1}}={{x}^{4}}$

Power of a Product$\displaystyle {{a}^{n}}\cdot {{b}^{n}}={{\left( a\cdot b \right)}^{n}}$, so to multiply two different bases with the same exponent, you need to multiply the bases and raise them to the common power.

Worked example 5:

$\displaystyle {{4}^{3}}\cdot {{5}^{3}}={{\left( 4\cdot 5 \right)}^{3}}$

$\displaystyle ={{20}^{3}}$

Quotient Rule: $\displaystyle \frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$, so, to divide two exponents with the same base, you keep the base and subtract the powers.

Worked example 6:

$\displaystyle \frac{{{5}^{5}}}{{{5}^{3}}}={{5}^{5-3}}={{5}^{2}}$

$\displaystyle \frac{{{{y}^{9}}}}{y}={{y}^{{9-1}}}={{y}^{8}}$

Exercise 1: Simplify by using the rules of exponents.

Check out order of operations BODMAS for recall.

a) $\displaystyle {{(-2)}^{3}}\div {{\left( {-\frac{1}{2}} \right)}^{{-2}}}+{{(-4)}^{2}}$

$\displaystyle {{(-2)}^{3}}\div {{\left( {-\frac{1}{2}} \right)}^{{-2}}}+{{(-4)}^{2}}=$$\displaystyle {{(-2)}^{3}}\div {{(-{{2}^{{-1}}})}^{{-2}}}+{{(-4)}^{2}}=$$ \displaystyle {{(-2)}^{3}}\div {{(-2)}^{2}}+{{(-4)}^{2}}=$$\displaystyle {{(-2)}^{{(3-2)}}}+{{(-4)}^{2}}=$$ \displaystyle {{(-2)}^{1}}+16=$$\displaystyle -2+16=14 a) \displaystyle {{(-2)}^{3}}\div {{\left( {-\frac{1}{2}} \right)}^{{-2}}}+{{(-4)}^{2}} \displaystyle {{(-2)}^{3}}\div {{\left( {-\frac{1}{2}} \right)}^{{-2}}}+{{(-4)}^{2}}= \displaystyle {{(-2)}^{3}}\div {{(-{{2}^{{-1}}})}^{{-2}}}+{{(-4)}^{2}}=$$ \displaystyle {{(-2)}^{3}}\div {{(-2)}^{2}}+{{(-4)}^{2}}=$

$\displaystyle {{(-2)}^{{(3-2)}}}+{{(-4)}^{2}}=$

$\displaystyle {{(-2)}^{1}}+16=$

$\displaystyle -2+16=14$

b) $\displaystyle \frac{{0.9\cdot {{{10}}^{{-3}}}+0.03\cdot {{{10}}^{{-2}}}}}{{1.2\cdot {{{10}}^{{-4}}}}}$

$\displaystyle \frac{{0.9\cdot {{{10}}^{{-3}}}+0.03\cdot {{{10}}^{{-2}}}}}{{1.2\cdot {{{10}}^{{-4}}}}}=$$\displaystyle \frac{{{{{10}}^{4}}(0.9\cdot {{{10}}^{{-3}}}+0.03\cdot {{{10}}^{{-2}}})}}{{1.2}}=$$ \displaystyle \frac{{0.9\cdot {{{10}}^{{(4-3)}}}+0.03\cdot {{{10}}^{{(4-2)}}}}}{{1.2}}=$$\displaystyle \frac{{0.9\cdot 10+0.03\cdot {{{10}}^{2}}}}{{1.2}}=$$ \displaystyle \frac{{9+3}}{{1.2}}=\frac{{12}}{{1.2}}=10$

b) $\displaystyle \frac{{0.9\cdot {{{10}}^{{-3}}}+0.03\cdot {{{10}}^{{-2}}}}}{{1.2\cdot {{{10}}^{{-4}}}}}$

$\displaystyle \frac{{0.9\cdot {{{10}}^{{-3}}}+0.03\cdot {{{10}}^{{-2}}}}}{{1.2\cdot {{{10}}^{{-4}}}}}=$

$\displaystyle \frac{{{{{10}}^{4}}(0.9\cdot {{{10}}^{{-3}}}+0.03\cdot {{{10}}^{{-2}}})}}{{1.2}}=$

$\displaystyle \frac{{0.9\cdot {{{10}}^{{(4-3)}}}+0.03\cdot {{{10}}^{{(4-2)}}}}}{{1.2}}=$

$\displaystyle \frac{{0.9\cdot 10+0.03\cdot {{{10}}^{2}}}}{{1.2}}=$

$\displaystyle \frac{{9+3}}{{1.2}}=\frac{{12}}{{1.2}}=10$

c) $\displaystyle \frac{{{{{10}}^{{-2}}}\cdot {{6}^{{-3}}}\cdot {{4}^{{-1}}}}}{{{{{12}}^{{-3}}}\cdot {{8}^{{-2}}}\cdot {{{10}}^{{-3}}}}}$

$\displaystyle \frac{{{{{10}}^{{-2}}}\cdot {{6}^{{-3}}}\cdot {{4}^{{-1}}}}}{{{{{12}}^{{-3}}}\cdot {{8}^{{-2}}}\cdot {{{10}}^{{-3}}}}}=$ $\displaystyle \frac{{{{{12}}^{3}}\cdot {{8}^{2}}\cdot {{{10}}^{3}}}}{{{{{10}}^{2}}\cdot {{6}^{3}}\cdot {{4}^{1}}}}=$$\displaystyle \frac{{{{{(2\cdot 6)}}^{3}}\cdot {{{(2\cdot 4)}}^{2}}\cdot {{{10}}^{3}}}}{{{{{10}}^{2}}\cdot {{6}^{3}}\cdot {{4}^{1}}}}=$$ \displaystyle \frac{{{{2}^{3}}\cdot {{6}^{3}}\cdot {{2}^{2}}\cdot {{4}^{2}}\cdot {{{10}}^{3}}}}{{{{{10}}^{2}}\cdot {{6}^{3}}\cdot {{4}^{1}}}}=$$\displaystyle \frac{{{{2}^{3}}\cdot \cancel{{{{6}^{3}}}}\cdot {{2}^{2}}\cdot {{4}^{2}}\cdot {{{10}}^{3}}}}{{{{{10}}^{2}}\cdot \cancel{{{{6}^{3}}}}\cdot {{4}^{1}}}}=$$ \displaystyle \frac{{8\cdot 4\cdot 16\cdot {{{10}}^{3}}}}{{{{{10}}^{2}}\cdot 4}}=$$\displaystyle \frac{{8\cdot \cancel{4}\cdot 16\cdot {{{10}}^{{3-2}}}}}{{\cancel{4}}}=1280 c) \displaystyle \frac{{{{{10}}^{{-2}}}\cdot {{6}^{{-3}}}\cdot {{4}^{{-1}}}}}{{{{{12}}^{{-3}}}\cdot {{8}^{{-2}}}\cdot {{{10}}^{{-3}}}}} \displaystyle \frac{{{{{10}}^{{-2}}}\cdot {{6}^{{-3}}}\cdot {{4}^{{-1}}}}}{{{{{12}}^{{-3}}}\cdot {{8}^{{-2}}}\cdot {{{10}}^{{-3}}}}}= \displaystyle \frac{{{{{12}}^{3}}\cdot {{8}^{2}}\cdot {{{10}}^{3}}}}{{{{{10}}^{2}}\cdot {{6}^{3}}\cdot {{4}^{1}}}}= \displaystyle \frac{{{{{(2\cdot 6)}}^{3}}\cdot {{{(2\cdot 4)}}^{2}}\cdot {{{10}}^{3}}}}{{{{{10}}^{2}}\cdot {{6}^{3}}\cdot {{4}^{1}}}}= \displaystyle \frac{{{{2}^{3}}\cdot {{6}^{3}}\cdot {{2}^{2}}\cdot {{4}^{2}}\cdot {{{10}}^{3}}}}{{{{{10}}^{2}}\cdot {{6}^{3}}\cdot {{4}^{1}}}}= \displaystyle \frac{{{{2}^{3}}\cdot \cancel{{{{6}^{3}}}}\cdot {{2}^{2}}\cdot {{4}^{2}}\cdot {{{10}}^{3}}}}{{{{{10}}^{2}}\cdot \cancel{{{{6}^{3}}}}\cdot {{4}^{1}}}}= \displaystyle \frac{{8\cdot 4\cdot 16\cdot {{{10}}^{3}}}}{{{{{10}}^{2}}\cdot 4}}= \displaystyle \frac{{8\cdot \cancel{4}\cdot 16\cdot {{{10}}^{{3-2}}}}}{{\cancel{4}}}=1280 d) \displaystyle \frac{{{{2}^{{2n-1}}}}}{{{{4}^{{n+1}}}}}\cdot \frac{{{{8}^{n}}}}{{{{2}^{{n-1}}}}} \displaystyle \frac{{{{2}^{{2n-1}}}}}{{{{4}^{{n+1}}}}}\cdot \frac{{{{8}^{n}}}}{{{{2}^{{n-1}}}}}=$$ \displaystyle \frac{{{{2}^{{2n-1}}}}}{{{{2}^{{2(n+1)}}}}}\cdot \frac{{{{2}^{{3n}}}}}{{{{2}^{{n-1}}}}}=$$\displaystyle \frac{{{{2}^{{(2n-1)+(3n)}}}}}{{{{2}^{{(2n+2)+(n-1)}}}}}=$$ \displaystyle \frac{{{{2}^{{5n-1}}}}}{{{{2}^{{3n+1}}}}}=$$\displaystyle {{2}^{{(5n-1)-(3n+1)}}}=$$ \displaystyle {{2}^{{5n-3n-1-1}}}=$$\displaystyle {{2}^{{2n-2}}} d) \displaystyle \frac{{{{2}^{{2n-1}}}}}{{{{4}^{{n+1}}}}}\cdot \frac{{{{8}^{n}}}}{{{{2}^{{n-1}}}}} \displaystyle \frac{{{{2}^{{2n-1}}}}}{{{{4}^{{n+1}}}}}\cdot \frac{{{{8}^{n}}}}{{{{2}^{{n-1}}}}}= \displaystyle \frac{{{{2}^{{2n-1}}}}}{{{{2}^{{2(n+1)}}}}}\cdot \frac{{{{2}^{{3n}}}}}{{{{2}^{{n-1}}}}}= \displaystyle \frac{{{{2}^{{(2n-1)+(3n)}}}}}{{{{2}^{{(2n+2)+(n-1)}}}}}=$$ \displaystyle \frac{{{{2}^{{5n-1}}}}}{{{{2}^{{3n+1}}}}}=$

$\displaystyle {{2}^{{(5n-1)-(3n+1)}}}=$

$\displaystyle {{2}^{{5n-3n-1-1}}}=$

$\displaystyle {{2}^{{2n-2}}}$

e) $\displaystyle \frac{{{{9}^{{\frac{a}{2}+1}}}-{{{27}}^{{\frac{a}{3}+1}}}}}{{{{{81}}^{{\frac{a}{4}}}}+{{3}^{a}}}}$

$\displaystyle \frac{{{{9}^{{\frac{a}{2}+1}}}-{{{27}}^{{\frac{a}{3}+1}}}}}{{{{{81}}^{{\frac{a}{4}}}}+{{3}^{a}}}}=$$\displaystyle \frac{{{{3}^{{2(\frac{a}{2}+1)}}}-{{3}^{{3(\frac{a}{3}+1)}}}}}{{{{3}^{{4(\frac{a}{4})}}}+{{3}^{a}}}}=$$ \displaystyle \frac{{{{3}^{{a+2}}}-{{3}^{{a+3}}}}}{{{{3}^{a}}+{{3}^{a}}}}=$$\displaystyle \frac{{{{3}^{a}}({{3}^{2}}-{{3}^{3}})}}{{2\cdot {{3}^{a}}}}=$$ \displaystyle \frac{{\cancel{{{{3}^{a}}}}(9-27)}}{{2\cdot \cancel{{{{3}^{a}}}}}}=$ $\displaystyle \frac{{18}}{2}=9$

e) $\displaystyle \frac{{{{9}^{{\frac{a}{2}+1}}}-{{{27}}^{{\frac{a}{3}+1}}}}}{{{{{81}}^{{\frac{a}{4}}}}+{{3}^{a}}}}$

$\displaystyle \frac{{{{9}^{{\frac{a}{2}+1}}}-{{{27}}^{{\frac{a}{3}+1}}}}}{{{{{81}}^{{\frac{a}{4}}}}+{{3}^{a}}}}=$

$\displaystyle \frac{{{{3}^{{2(\frac{a}{2}+1)}}}-{{3}^{{3(\frac{a}{3}+1)}}}}}{{{{3}^{{4(\frac{a}{4})}}}+{{3}^{a}}}}=$

$\displaystyle \frac{{{{3}^{{a+2}}}-{{3}^{{a+3}}}}}{{{{3}^{a}}+{{3}^{a}}}}=$

$\displaystyle \frac{{{{3}^{a}}({{3}^{2}}-{{3}^{3}})}}{{2\cdot {{3}^{a}}}}=$

$\displaystyle \frac{{\cancel{{{{3}^{a}}}}(9-27)}}{{2\cdot \cancel{{{{3}^{a}}}}}}=$

$\displaystyle \frac{{18}}{2}=9$

f) $\displaystyle \frac{{{{2}^{{2006}}}-{{2}^{{2005}}}}}{{2\cdot {{2}^{{2002}}}}}$

$\displaystyle \frac{{{{2}^{{2006}}}-{{2}^{{2005}}}}}{{2\cdot {{2}^{{2002}}}}}=$$\displaystyle \frac{{{{2}^{{2005}}}\cdot {{2}^{1}}-{{2}^{{2005}}}}}{{{{2}^{{1+2002}}}}}=$$ \displaystyle \frac{{{{2}^{{2005}}}(2-1)}}{{{{2}^{{2003}}}}}=$$\displaystyle \frac{{{{2}^{{2005}}}}}{{{{2}^{{2003}}}}}=$$ \displaystyle {{2}^{{(2005-2003)}}}=$$\displaystyle {{2}^{2}}=4$

f) $\displaystyle \frac{{{{2}^{{2006}}}-{{2}^{{2005}}}}}{{2\cdot {{2}^{{2002}}}}}$

$\displaystyle \frac{{{{2}^{{2006}}}-{{2}^{{2005}}}}}{{2\cdot {{2}^{{2002}}}}}=$

$\displaystyle \frac{{{{2}^{{2005}}}\cdot {{2}^{1}}-{{2}^{{2005}}}}}{{{{2}^{{1+2002}}}}}=$

$\displaystyle \frac{{{{2}^{{2005}}}(2-1)}}{{{{2}^{{2003}}}}}=$

$\displaystyle \frac{{{{2}^{{2005}}}}}{{{{2}^{{2003}}}}}=$

$\displaystyle {{2}^{{(2005-2003)}}}=$

$\displaystyle {{2}^{2}}=4$

Exercise 2: Find x.

a) $\displaystyle \frac{{{{4}^{x}}+{{4}^{x}}+{{4}^{x}}+{{4}^{x}}}}{{{{2}^{x}}+{{2}^{x}}}}=32$

$\displaystyle \frac{{4\cdot {{4}^{x}}}}{{2\cdot {{2}^{x}}}}=32$

$\displaystyle \frac{{\cancel{4}\cdot {{4}^{x}}}}{{\cancel{2}\cdot {{2}^{x}}}}=32$

$\displaystyle \frac{{2\cdot {{2}^{2}}^{x}}}{{{{2}^{x}}}}={{2}^{5}}$

$\displaystyle \frac{{{{2}^{{1+2x}}}}}{{{{2}^{x}}}}={{2}^{5}}$

$\displaystyle {{2}^{{1+2x-x}}}={{2}^{5}}$

$\displaystyle {{2}^{{1+x}}}={{2}^{5}}$

$\displaystyle 1+x=5$

$\displaystyle x=5-1=4$

b) $\displaystyle \frac{{{{6}^{x}}+{{9}^{x}}}}{{{{4}^{x}}+{{6}^{x}}}}=\frac{8}{{27}}$

$\displaystyle \frac{{{{{(2\cdot 3)}}^{x}}+{{{(3\cdot 3)}}^{x}}}}{{{{{(2\cdot 2)}}^{x}}+{{{(2\cdot 3)}}^{x}}}}=\frac{8}{{27}}$

$\displaystyle \frac{{{{2}^{x}}\cdot {{3}^{x}}+{{3}^{x}}\cdot {{3}^{x}}}}{{{{2}^{x}}\cdot {{2}^{x}}+{{2}^{x}}\cdot {{3}^{x}}}}=\frac{8}{{27}}$

$\displaystyle \frac{{{{3}^{x}}({{2}^{x}}+{{3}^{x}})}}{{{{2}^{x}}({{2}^{x}}+{{3}^{x}})}}=\frac{8}{{27}}$

$\displaystyle \frac{{{{3}^{x}}\cancel{{({{2}^{x}}+{{3}^{x}})}}}}{{{{2}^{x}}\cancel{{({{2}^{x}}+{{3}^{x}})}}}}=\frac{8}{{27}}$

$\displaystyle \frac{{{{3}^{x}}}}{{{{2}^{x}}}}=\frac{{{{2}^{3}}}}{{{{3}^{3}}}}$

$\displaystyle {{\left( {\frac{3}{2}} \right)}^{x}}={{\left( {\frac{2}{3}} \right)}^{3}}$

$\displaystyle {{\left( {\frac{3}{2}} \right)}^{x}}={{\left( {\frac{3}{2}} \right)}^{{-3}}}$

$\displaystyle x=-3$

c) $\displaystyle {{\left( {\frac{4}{9}} \right)}^{x}}\cdot {{\left( {\frac{{27}}{8}} \right)}^{{2x-1}}}=\left( {\frac{3}{2}} \right)$

$\displaystyle {{\left( {\frac{{{{2}^{2}}}}{{{{3}^{2}}}}} \right)}^{x}}\cdot {{\left( {\frac{{{{3}^{3}}}}{{{{2}^{3}}}}} \right)}^{{2x-1}}}=\left( {\frac{3}{2}} \right)$

$\displaystyle {{\left( {\frac{2}{3}} \right)}^{{2x}}}\cdot {{\left( {\frac{3}{2}} \right)}^{{3(2x-1)}}}=\left( {\frac{3}{2}} \right)$

$\displaystyle {{\left( {\frac{2}{3}} \right)}^{{2x}}}\cdot {{\left( {\frac{2}{3}} \right)}^{{-3(2x-1)}}}={{\left( {\frac{2}{3}} \right)}^{{-1}}}$

$\displaystyle {{\left( {\frac{2}{3}} \right)}^{{2x}}}\cdot {{\left( {\frac{2}{3}} \right)}^{{-6x+3}}}={{\left( {\frac{2}{3}} \right)}^{{-1}}}$

$\displaystyle 2x+3-6x=-1$

$\displaystyle 3-4x=-1$

$\displaystyle -4x=-4$

$\displaystyle x=1$

d) $\displaystyle {{3}^{x}}+{{3}^{{x+1}}}+{{3}^{{x+2}}}={{5}^{x}}+{{5}^{{x+1}}}+{{5}^{{x+2}}}$

$\displaystyle {{3}^{x}}+{{3}^{x}}\cdot {{3}^{1}}+{{3}^{x}}\cdot {{3}^{2}}={{5}^{x}}+{{5}^{x}}\cdot {{5}^{1}}+{{5}^{x}}\cdot {{5}^{2}}$

$\displaystyle {{3}^{x}}(1+3+{{3}^{2}})={{5}^{x}}(1+5+{{5}^{2}})$

$\displaystyle 13\cdot {{3}^{x}}=31\cdot {{5}^{x}}$

$\displaystyle \frac{{{{3}^{x}}}}{{{{5}^{x}}}}=\frac{{31}}{{13}}$

$\displaystyle {{\left( {\frac{3}{5}} \right)}^{x}}=\frac{{31}}{{13}}$

$\displaystyle x={{\log }_{{\frac{3}{5}}}}\frac{{31}}{{13}}$