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Calculating with Fractions

Solution ("long way").

We multiply straight across, as follows:

\( \dfrac{7\cdot 5}{3\cdot 21}\)

which becomes

\( \dfrac{35}{63} \)

Next, we divide out a 7 from numerator and denominator:

\( \dfrac{35\color{red}\div 7}{63\color{red}\div 7} \)

which simplifies to

\( \dfrac{5}{9} \)
Solution ("shortcut").

We cancel a pair of 7's, as follows:

\( \dfrac{\cancel{7}}{3}\cdot\dfrac{5}{\cancel{21}\;\;^{\color{red}3}} \)

Then we multiply across:

\( \dfrac{5}{3\cdot 3} \)

which simplifies to

\( \dfrac{5}{9} \)

click for video explanation

Solution ("long way").

We begin by converting the whole number into a fraction (over \(1\)):

\( \dfrac{7}{6}\cdot\dfrac{8}{\color{red}1} \)

Next, we multiply straight across:

\( \dfrac{7\cdot 8}{6\cdot 1}\)

which becomes

\( \dfrac{56}{6} \)

Next, we divide out a 2 from numerator and denominator:

\( \dfrac{56\color{red}\div 2}{6\color{red}\div 2} \)

which simplifies to

\( \dfrac{28}{3} \)
Solution ("shortcut").

We cancel out a 2 from the 6 and 4, as follows:

\( \dfrac{7}{^{\color{red}3}\;\;\cancel{6}}\cdot \cancel{8}\;\;^{\color{red}4} \)



Then we multiply the whole number 4 into the numerator:

\( \dfrac{7\cdot 4}{3} \)



which simplifies to
\( \dfrac{28}{3} \)

click for video explanation

Solution. We begin by flipping the divisor to convert to multiplication:

\( \dfrac{10}{9}\cdot\dfrac{2}{5}\)

Then we multiply straight across

\( \dfrac{10\cdot 2}{9\cdot 5} \)

which becomes

\( \dfrac{20}{45} \)

Next, we divide out a 5 from numerator and denominator:

\( \dfrac{20\color{red}\div 5}{45\color{red}\div 5} \)

which simplifies to

\( \dfrac{4}{9} \)

click for video explanation

Solution. When dividing a fraction by a whole number, we have to first convert the whole number to a fraction (over \(1\)):

\( \dfrac{21}{8}\div\dfrac{14}{\color{red}1} \)

Division then turns into multiplication of the reciprocal:

\( \dfrac{21}{8}\color{red}\cdot\dfrac{1}{14} \)

We can the multiply straight across:

\( \dfrac{21\cdot 1}{8\cdot 14}\)

which becomes

\( \dfrac{21}{112} \)

Next, we divide out a 7 from numerator and denominator:

\( \dfrac{21\color{red}\div 7}{112\color{red}\div 7} \)

which simplifies to

\( \dfrac{3}{16} \)

click for video explanation

Solution. First we rewrite \(15\) as \(15/1\):

\( \dfrac{15}{\color{red}1}\div\dfrac{5}{3} \)

Next, we convert division into multiplication by the reciprocal:

\( \dfrac{15}{1}\color{red}\cdot\dfrac{3}{5} \)

Then we multiply across:

\( \dfrac{15\cdot 3}{1\cdot 5} \)

which simplifies to:

\( \dfrac{45}{5} \)

and hence:

\( 9 \)

click for video explanation

Solution. We begin by converting division into multiplication by the reciprocal:

\( \dfrac{4}{9}\color{red}\cdot\dfrac{21}{10} \)

Next, we multiply straight across:

\( \dfrac{4\cdot 21}{9\cdot 10} \)

which simplifies to:

\( \dfrac{84}{90} \)

We can now divide out \(6\) from numerator and denominator:

\( \dfrac{84\color{red}\div 6}{90\color{red}\div 6} \)

which simplifies to:

\( \dfrac{14}{15} \)

click for video explanation

Solution. We begin by converting \(12\) to \(12/1\):

\( \dfrac{\textstyle\frac{4}{3}}{\textstyle\frac{12}{\color{red}1}} \)

Division now turns into multiplication by the reciprocal:

\( \dfrac{4}{3}\color{red}\cdot\dfrac{1}{12} \)

Next, we multiply straight across:

\( \dfrac{4\cdot 1}{3\cdot 12} \)

which simplifies to:

\( \dfrac{4}{36} \)

Finally, we can divide out a \(4\) from numerator and denominator:

\( \dfrac{4\color{red}\div 4}{36\color{red}\div 4} \)

which simplifies to:

\( \dfrac{1}{9} \)

click for video explanation

Solution. We begin by rewriting \(11\) as \(11/1\):

\( \dfrac{\textstyle\frac{11}{\color{red}1}}{\textstyle\frac{44}{9}} \)

Next, division turns into multiplication by the reciprocal:

\( \dfrac{11}{1}\color{red}\cdot\dfrac{9}{44} \)

We can now multiply straight across:

\( \dfrac{11\cdot 9}{1\cdot 44} \)

which simplifies to:

\( \dfrac{99}{44} \)

And finally we can divide out \(11\) from numerator and denominator:

\( \dfrac{99\color{red}\div 11}{44\color{red}\div 11} \)

which simplifies to:

\( \dfrac{9}{4} \)

click for video explanation

Solution. We begin by multiplying the first fraction by 7/7 and the second fraction by 3/3:

\( \dfrac{2\color{red}\cdot 7}{3\color{red}\cdot 7}+\dfrac{5\color{red}\cdot 3}{7\color{red}\cdot 3}\)

which gives us the common denominators

\( \dfrac{14}{21}+\dfrac{15}{21} \)

Next, we add the numerators (and keep the common denominator):

\( \dfrac{14+15}{21} \)

which simplifies to

\( \dfrac{29}{21} \)

click for video explanation

Solution. We begin by rewriting the whole number 3 as a fraction 3/1, as follows:

\( \dfrac{3}{\color{red}1}+\dfrac{9}{13}\)

Then we multiply the first fraction by 13/13:

\( \dfrac{3\color{red}\cdot 13}{1\color{red}\cdot 13}+\dfrac{9}{13} \)

which gives us common denominators

\( \dfrac{39}{13}+\dfrac{9}{13} \)

Next, we add the numerators and keep the common denominator:

\( \dfrac{39+9}{13} \)

which simplifies to

\( \dfrac{48}{13} \)

click for video explanation

Solution. We begin by multiplying the first fraction by 5/5 and the second by 3/3, as follows:

\( \dfrac{\color{red}5\cdot\color{black}5}{\color{red}5\cdot \color{black}3}-\dfrac{12\color{red}\cdot 3}{5\color{red}\cdot 3}\)

which gives us common denominators:

\( \dfrac{25}{15}-\dfrac{36}{15} \)

Next we subtract the numerators (keeping the common denominator):

\( \dfrac{25-36}{15} \)

which simplifies to

\( -\dfrac{11}{15} \)

click for video explanation

Solution. We begin by rewriting the whole number 7 as the fraction 7/1:

\( \dfrac{7}{\color{red}1}-\dfrac{5}{11}\)

Next, we multiply the first fraction by 11/11:

\( \dfrac{\color{red}11\cdot\color{black}7}{\color{red}11\cdot\color{black}1}-\dfrac{5}{11} \)

which gives us the common denominators

\( \dfrac{77}{11}-\dfrac{5}{11} \)

Then we subtract the numerators (keeping the common denominator):

\( \dfrac{77-5}{11} \)

which simplifies to

\( \dfrac{72}{11} \)

click for video explanation

Factoring

Solution. Remember that the first few prime numbers are:

\(2,\; 3,\; 5,\; 7,\; 11,\; 13,\; 17,\; 19,\; 23,\; 29,\; \cdots \)

We're going to try each of these until we're finished. Note that

\( 78\div 2 = 39\)

and that

\( 39\div 3 = 13 \)

Hence, the prime factorization is:

\( 2\cdot 3\cdot 13 \)

click for video explanation

Solution. Remember that the first few prime numbers are:

\(2,\; 3,\; 5,\; 7,\; 11,\; 13,\; 17,\; 19,\; 23,\; 29,\; \cdots \)

We're going to try each of these until we're finished. Note that

\( 1960\div 2 = 980\)

and that

\( 980\div 2 = 490 \)

and also

\( 490\div 2 = 245 \)

Then we try:

\( 245\div 5 = 49 \)

and

\( 49\div 7 = 7 \)

Hence, the prime factorization is

\( 2\cdot 2\cdot 2\cdot 5\cdot 7\cdot 7 \)

which we rewrite as

\( 2^3\cdot 5\cdot 7^2 \)

click for video explanation

Integer Exponents

Whole number exponents. A whole number, positive exponent \(n\), takes its base \(b\) and multiplies it with itself \(n\) times:

\(b^n=\underbrace{b\cdot b\cdot b\cdot \cdots \cdot b}_{n\text{ times}}\)

Example.
\( b^5= b\cdot b\cdot b\cdot b\cdot b\)

Example.
\( \begin{align*} 3^5 &=3\cdot 3\cdot 3\cdot 3\cdot 3 \\\\ &=243. \end{align*} \)

click for video explanation


Negative bases. A negative is applied after the exponent, unless parentheses force the exponent to be applied first.

Example.
\( (-3)^4=81 \)

Example.
\( -3^4=-81 \)

click for video explanation


The zero exponent. Any real number raised to the zeroth power is equal to \(1\), except for zero itself. In other words:

\( b^0=1 \;\;\;(\text{if } b\neq 0) \)

Example.
\( 5^0=1 \)

Example.
\( (-3)^0=1 \)

Example.
\( -3^0=-1 \)

Example.
\( 0^0 \;\text{ is undefined.} \)

click for video explanation


Negative exponents. Negative exponents are defined via the rule

\( b^{-n}=\dfrac{1}{b^n} \)

Example.
\( 4^{-3}=\dfrac{1}{4^3}=\dfrac{1}{64} \)

Example.
\( 5^{-1}=\dfrac{1}{5} \)

Example.
\( (-2)^{-5}=\dfrac{1}{(-2)^5}=\dfrac{1}{-32} \)

click for video explanation

Laws of exponents.

1. \( b^0=1 \)

2. \( b^1=b \)

3. \( b^mb^n=b^{m+n} \)

4. \( \displaystyle\frac{b^m}{b^n}=b^{m-n} \;\;\; \) or \( \;\;\; \displaystyle\frac{b^m}{b^n}=\frac{1}{b^{n-m}} \)

5. \( (ab)^n=a^nb^n \)

6. \( \displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n} \)

7. \( \displaystyle\left(b^m\right)^n=b^{mn} \)

8. \( \displaystyle b^{-n}=\frac{1}{b^n} \)

9. \( \displaystyle\left(\frac{a}{b}\right)^{-n}=\frac{b^n}{a^n} \)

10. \( \displaystyle b^{1/n}=\sqrt[n]{b} \)

11. \( \displaystyle b^{m/n}=\left(\sqrt[n]{b}\right)^m \;\;\; \) or \( \;\;\; \displaystyle b^{m/n}=\sqrt[n]{b^m}\)

Note: If \( b\in(0,1)\cup(1,\infty) \) then the one-to-one property holds:

\( b^m=b^n\;\;\;\) if and only if \( \;\;\;m=n \).

Solution.
\( (3x)^0=1 \)

Solution.
\( (2z)^1=2z \)

Solution.
\( x^4x^9=x^{\color{red}4+9}=x^{13} \)

Solution.
\( \dfrac{y^8}{y^3}=y^{\color{red}8-3}=y^5 \)

Solution.
\( \dfrac{z^3}{z^9}=\dfrac{1}{z^{\color{red}9-3}}=\dfrac{1}{z^6} \)

Solution.
\( (2xy)^3=2^{\color{red}3}x^{\color{red}3}y^{\color{red}3}=8x^3y^3 \)

Solution.
\( \left(\dfrac{5x}{y}\right)^2=\dfrac{5^{\color{red}2}x^{\color{red}2}}{y^{\color{red}2}}=\dfrac{25x^2}{y^2} \)

Solution.
\( (x^3)^5=x^{\color{red}3\cdot 5}=x^{15} \)

Solution. We have
\( \begin{align*} \displaystyle\left(7x^{-5}y^4\right)^2 &=\color{red}49\color{black}x^{\color{red}-10}y^{\color{red}8} \\\\ &=\dfrac{49y^8}{x^{10}}. \end{align*} \)

click for video explanation

Solution. We have
\( \begin{align*} \displaystyle\left(-3y^3z^2\right)^3\left(y^2z\right)^{-4} &=\left(\color{red}-27\color{black}y^{\color{red}9}z^{\color{red}6}\right)\left(y^{\color{red}-8}z^{\color{red}-4}\right) \\\\ &=-27yz^2. \end{align*} \)

click for video explanation

Solution. We have
\( \begin{align*} \dfrac{x^{-3}y}{8x^{-4}y^4} &=\dfrac{\color{red}x^4\color{black}y}{8\color{red}x^3\color{black}y^4} \\\\ &=\dfrac{x}{8y^3}. \end{align*} \)

click for video explanation

Solution. We have
\( \begin{align*} \dfrac{7a^{-1}b^3}{\displaystyle\left(3b^4\right)^{-2}a^{17}} &=\dfrac{7a^{-1}b^3}{3^{\color{red}-2}b^{\color{red}-8}a^{17}} \\\\ &=\dfrac{7\cdot\color{red}3^2\color{black}b^3\color{red}b^8}{\color{red}a\color{black}\cdot a^{17}} \\\\ &=\dfrac{63b^{11}}{a^{18}}. \end{align*} \)

click for video explanation

Solution. We have
\( \begin{align*} \left(\dfrac{w^{-4}z^5}{z^{-5}w^{-1}}\right)^7 &=\left(\dfrac{\color{red}z^5\color{black}z^5\color{red}w}{\color{red}w^4}\right)^7 \\\\ &=\left(\dfrac{\color{red}z^{10}}{\color{red}w^3}\right)^7 \\\\ &=\dfrac{z^{70}}{w^{21}}. \end{align*} \)

click for video explanation

Solution. We have
\( \begin{align*} \left(\dfrac{5x^3y^{-7}}{7x^{-4}y}\right)^{-3} &=\left(\dfrac{5x^3\color{red}x^4}{7y\cdot\color{red}y^7}\right)^{-3} \\\\ &=\left(\dfrac{5x^7}{7y^8}\right)^{-3} \\\\ &=\left(\dfrac{\color{blue}7y^8}{\color{blue}5x^7}\right)^{\color{red}3} \\\\ &=\dfrac{343y^{24}}{125x^{21}}. \end{align*} \)

click for video explanation

Logarithms

Logarithmic and exponential forms.

\( \underbrace{y=\log_b(x)}_{\text{logarthmic form}}\;\;\; \) if and only if \( \;\;\;\underbrace{b^y=x}_{\text{exponential form}} \)

Here, \(b\) is called the base and \(y\) is called the exponent.

Example. The following is in logarithmic form:

\( {\color{red}2}=\log_{\color{blue}3}({\color{Plum}9}) \)

In exponential form, this becomes:

\( {\color{blue}3}^{\color{red}2}={\color{Plum}9} \)

Example. It doesn't matter if we write left-to-right or right-to-left. Indeed, the following is in logarithmic form:

\( \log_{\color{blue}1/2}({\color{Plum}32})={\color{red}-5} \)

In exponential form, this becomes:

\( {\color{Plum}32}=\left({\color{blue}\dfrac{1}{2}}\right)^{\color{red}-5} \)

click for video explanation


Common and natural logarithms, and parentheses.

The common logarithm is base 10, and has the following abbreviation:

"\(\log\)"\(\;\;\;\;\) means \(\;\;\;\;\)"\(\log_{10}\)"

Recall that Euler's number has the approximation

\( e\approx 2.718\;281\;828\;459\;045\cdots \)

The natural logarithm is base \(e\), and has the following abbreviation:

"\(\ln\)"\(\;\;\;\;\) means \(\;\;\;\;\)"\(\log_e\)"

Example. The exponential form \({\color{blue}10}^{\color{red}3}={\color{Plum}1000}\) can be converted to logarithmic form:

\(\log{\color{Plum}1000}={\color{red}3}\)

Example. The exponential form \(e^1=e\) can be converted to logarithmic form:

\(\ln{\color{Plum}e}={\color{red}1}\)

Note: We may suppress the parentheses when writing logarithms with inputs that are a constant or single variable. For instance,

\(\log_b(m)=\log_b m\).

However, be aware that this is not the case all the time:

\(\log_b(x+1)\neq\log_b x+1\).

click for video explanation

Laws of logarithms.

1. \( \log_b 1=0 \)

2. \( \log_b b=1 \)

3. \( \log_b b^p=p \)

4. \( b^{\log_b m}=m \)

5. \( \log_b mn=\log_b m+\log_b n \)

6. \( \log_b\dfrac{m}{n}=\log_b m-\log_b n \)

7. \( \log_b m^p=p\log_b m \)

8. \( \log_b m=\dfrac{\log_a m}{\log_a b} \;\;\;\;\;\;\;\text{(change of base)}\)

Note: Additionally, the one-to-one property holds:

\( \log_bm=\log_bn\;\;\;\) if and only if \( \;\;\;m=n \).

Solution. We have:
\( \log_3 243=5 \)

click for video explanation

Solution. We have
\( \log_{\frac{1}{5}}25=-2 \)

click for video explanation

Solution. We have:
\( 7^3=343 \)

click for video explanation

Solution. We have:
\( 128^{-\frac{1}{7}}=\dfrac{1}{2} \)

click for video explanation

Solution. Recall that "\(\ln\)" means "\(\log_e\)", and so we have:

\( e^{-1}=\dfrac{1}{e} \)

click for video explanation

Solution. Converting to exponential form, we have:

\( {\color{red}b}^4=81 \)

Note that
\( {\color{red}3}^4=81 \)

Hence,
\( b=3 \)

click for video explanation

Solution. Using the laws of logarithms (#4), note that in

\( {\color{red}4}^{{\color{red}\log_4}7} \)

the \(4\) and \(\log_4\) cancel, simplifying to:

\(7\)

click for video explanation

Solution. Using the laws of logarithms (#3), note that in

\( {\color{red}\log_7}({\color{red}7}^9) \)

the \(\log_7\) and \(7\) cancel, simplifying to:

\(9\)

click for video explanation

Solution. Using the laws of logarithms (#5), this expands to:

\( \log_3(4)+\log_3(x) \)

click for video explanation

Solution. Using the laws of logarithms (#6), this expands to:

\( \log_7(y)-\log_7(x) \)

click for video explanation

Solution. We first subtract \(18\) from both sides:

\( \begin{align*} 18 & -7\log_3 x & = & -10 \\ \color{red}-18 & & & \color{red}-18 \\\hline & -7\log_3 x & = & -28 \end{align*} \)

Then we divide both sides by \(-7\):

\( \dfrac{-7\log_3 x}{\color{red}-7}=\dfrac{-28}{\color{red}-7} \)

which simplifies to:

\( \log_{\color{blue}3}{\color{plum}x}={\color{red}4} \)

This can be converted to exponential form as follows:

\({\color{blue}3}^{\color{red}4}={\color{plum}x}\)

And finally, this simplifies to:

\(x=81\)

Since this is a logarithmic equation, we must check our answer to make sure it is not extraneous:

\( 18-7\log_3{\color{red}81}=-10 \)

Since this statement is true (as verified in the calculator), it means that the answer is not extraneous. So the solution is:

\(x=81\)

click for video explanation

Solution. Using the laws of logarithms (#5, 6, and 7), this expands to:

\( \log_3(4)+7\log_3(x)-\log_3(5)-12\log_3(y) \)

click for video explanation

Solution. We first rewrite the square root as the \(1/2\) power:

\( \log_4\dfrac{5(x+1)^{\color{red}1/2}}{y} \)

Then we can use the laws of logarithms (#5, 6, and 7), to expand:

\( \log_4(5)+\dfrac{1}{2}\log_4(x+1)-\log_4(y) \)

click for video explanation

Solution. First, we distribute the cube root (as a \(1/3\) power):

\( \log_7\dfrac{8^{1\color{red}/3}(3x-2)^{6\color{red}/3}}{(y+4)^{1\color{red}/3}}, \)

which simplifies to

\( \log_7\dfrac{{\color{red}2}(3x-2)^{\color{red}2}}{(y+4)^{1/3}}. \)

Now we can expand:

\( \log_7 2 + 2\log_7(3x-2)-\dfrac{1}{3}\log_7(y+4) \)

click for video explanation

Solution. Note that
\( \begin{align*} \log_{11}\left(\dfrac{2}{9}\right) &=\log_{11}\left(\dfrac{2}{\color{red}3^2}\right) \\\\ &=\log_{11}(2)-2\log_{11}(3) \\\\ &\approx {\color{red}0.2891}-2({\color{red}0.4582}) \\\\ &=0.2891-\color{red}0.9164 \\\\ &=-0.6273. \end{align*} \)

click for video explanation