Use the Properties of Logarithms

Use the Properties of Logarithms

Now that we have learned about exponential and logarithmic functions, we can introduce some of the properties of logarithms. These will be very helpful as we continue to solve both exponential and logarithmic equations.

The first two properties derive from the definition of logarithms. Since \({a}^{0}=1,\) we can convert this to logarithmic form and get \({\text{log}}_{a}1=0.\) Also, since \({a}^{1}=a,\) we get \({\text{log}}_{a}a=1.\)

In the next example we could evaluate the logarithm by converting to exponential form, as we have done previously, but recognizing and then applying the properties saves time.

Example

Try it.

Evaluate using the properties of logarithms: ⓐ \({\text{log}}_{8}1\) and ⓑ \({\text{log}}_{6}6.\)

Solution

ⓐ

	\(\ {\text{log}}_{8}1\)
Use the property, \({\text{log}}_{a}1=0\).	\(\ 0\ {\text{log}}_{8}1=0\)

ⓑ
\(\begin{array}{llllll} & & & \ {\text{log}}_{6}6 & & \\ \text{Use the property,}\ {\text{log}}_{a}a=1. & & & \ 1 & & \ {\text{log}}_{6}6=1\end{array}\)

The next two properties can also be verified by converting them from exponential form to logarithmic form, or the reverse.

The exponential equation \({a}^{{\text{log}}_{a}x}=x\) converts to the logarithmic equation \({\text{log}}_{a}x={\text{log}}_{a}x,\) which is a true statement for positive values for x only.

The logarithmic equation \({\text{log}}_{a}{a}^{x}=x\) converts to the exponential equation \({a}^{x}={a}^{x},\) which is also a true statement.

These two properties are called inverse properties because, when we have the same base, raising to a power “undoes” the log and taking the log “undoes” raising to a power. These two properties show the composition of functions. Both ended up with the identity function which shows again that the exponential and logarithmic functions are inverse functions.

Example

Try it.

Evaluate using the properties of logarithms: ⓐ \({4}^{{\text{log}}_{4}9}\) and ⓑ \({\text{log}}_{3}{3}^{5}.\)

Solution

ⓐ

	\(\ {4}^{{\text{log}}_{4}9}\)
Use the property, \({a}^{{\text{log}}_{a}x}=x\).	\(\ 9\ {4}^{{\text{log}}_{4}9}=9\)

ⓑ

	\(\ {\text{log}}_{3}{3}^{5}\)
Use the property, \({a}^{{\text{log}}_{a}x}=x\).	\(\ 5\ {\text{log}}_{3}{3}^{5}=5\)

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Use the Change-of-Base Formula

To evaluate a logarithm with any other base, we can use the Change-of-Base Formula. We will show how this is derived.

Suppose we want to evaluate \({\text{log}}_{a}M\).	\(\ {\text{log}}_{a}M\\)
Let \(y={\text{log}}_{a}M\).	\(\ y\ =\ {\text{log}}_{a}M\)
Rewrite the expression in exponential form.	\(\ {a}^{y}\ =\ M\\)
Take the \({\text{log}}_{b}\) of each side.	\(\ {\text{log}}_{b}{a}^{y}\ =\ {\text{log}}_{b}M\)
Use the Power Property.	\(\ y{\text{log}}_{b}a\ =\ {\text{log}}_{b}M\)
Solve for \(y\).	\(\ y\ =\ \frac{{\text{log}}_{b}M}{{\text{log}}_{b}a}\)
Substitute \(y={\text{log}}_{a}M\).	\(\ {\text{log}}_{a}M\ =\ \frac{{\text{log}}_{b}M}{{\text{log}}_{b}a}\)

The Change-of-Base Formula introduces a new base \(b.\) This can be any base b we want where \(b>0,b\ne 1.\) Because our calculators have keys for logarithms base 10 and base e, we will rewrite the Change-of-Base Formula with the new base as 10 or e.

When we use a calculator to find the logarithm value, we usually round to three decimal places. This gives us an approximate value and so we use the approximately equal symbol \(\text{(\approx )}\).

Example

Try it.

Rounding to three decimal places, approximate \({\text{log}}_{4}35.\)

Solution

Use the Change-of-Base Formula.
Identify a and M. Choose 10 for b.
Enter the expression \(\frac{\text{log}35}{\text{log}4}\) in the calculator using the log button for base 10. Round to three decimal places.

Key Concepts

• Properties of Logarithms
  \[\ {\text{log}}_{a}1=0\ {\text{log}}_{a}a=1\]
• Inverse Properties of Logarithms
  • For \(a>0,\)\(x>0\) and \(a\ne 1\)
    \[{a}^{{\text{log}}_{a}x}=x\ {\text{log}}_{a}{a}^{x}=x\]
• Product Property of Logarithms
  
  • If \(M>0,N>0\text{,}\ \text{a}>0\) and \(\text{a}\ne 1,\) then,
    \[\ {\text{log}}_{a}M\cdot N={\text{log}}_{a}M+{\text{log}}_{a}N\]
    The logarithm of a product is the sum of the logarithms.
• Quotient Property of Logarithms
  
  • If \(M>0,N>0\text{,}\ \text{a}>0\) and \(\text{a}\ne 1,\) then,
    \[\ {\text{log}}_{a}\frac{M}{N}={\text{log}}_{a}M-{\text{log}}_{a}N\]
    The logarithm of a quotient is the difference of the logarithms.
• Power Property of Logarithms
  
  • If \(M>0,\ \text{a}>0,\ \text{a}\ne 1\) and \(p\) is any real number then,
    \[{\text{log}}_{a}{M}^{p}=p\ {\text{log}}_{a}M\]
    The log of a number raised to a power is the product of the power times the log of the number.
• Properties of Logarithms Summary
  If \(M>0,\ \text{a}>0,\ \text{a}\ne 1\) and \(p\) is any real number then,
  

  Property	Base \(a\)	Base \(e\)
  	\({\text{log}}_{a}1=0\)	\(\text{ln}1=0\)
  	\({\text{log}}_{a}a=1\)	\(\text{ln}\ e=1\)
  Inverse Properties	\(\begin{array}{l} \\ \\ {a}^{{\text{log}}_{a}x}=x \\ {\text{log}}_{a}{a}^{x}=x\end{array}\)	\(\begin{array}{l} \\ \\ {e}^{\text{ln}\ x}=x \\ \text{ln}\ {e}^{x}=x\end{array}\)
  Product Property of Logarithms	\({\text{log}}_{a}(M\cdot N)={\text{log}}_{a}M+{\text{log}}_{a}N\)	\(\text{ln}(M\ \cdot \ N)=\text{ln}\ M+\text{ln}\ N\)
  Quotient Property of Logarithms	\(\ {\text{log}}_{a}\frac{M}{N}={\text{log}}_{a}M-{\text{log}}_{a}N\)	\(\ \text{ln}\frac{M}{N}=\text{ln}\ M-\text{ln}\ N\)
  Power Property of Logarithms	\(\ {\text{log}}_{a}{M}^{p}=p\ {\text{log}}_{a}M\)	\(\ \text{ln}\ {M}^{p}=p\text{ln}\ M\)

• Change-of-Base Formula
  For any logarithmic bases a and b, and \(M>0,\)
  \[\begin{array}{lllllll}{\text{log}}_{a}M=\frac{{\text{log}}_{b}M}{{\text{log}}_{b}a} & & & \ {\text{log}}_{a}M=\frac{\text{log}M}{\text{log}a} & & & \ {\text{log}}_{a}M=\frac{\text{ln}\ M}{\text{ln}\ a} \\ \text{new base}\ b & & & \ \text{new base 10} & & & \ \text{new base}\ e\end{array}\]

Use the Properties of Logarithms

Use the Properties of Logarithms

In the following exercises, use the properties of logarithms to evaluate.

Try it.

ⓐ \({\text{log}}_{4}1\) ⓑ \({\text{log}}_{8}8\)

Try it.

ⓐ \({\text{log}}_{12}1\) ⓑ \(\text{ln}\ e\)

Solution

ⓐ 0 ⓑ 1

Try it.

ⓐ \({3}^{{\text{log}}_{3}6}\) ⓑ \({\text{log}}_{2}{2}^{7}\)

Try it.

ⓐ \({5}^{{\text{log}}_{5}10}\) ⓑ \({\text{log}}_{4}{4}^{10}\)

Solution

ⓐ 10 ⓑ 10

Try it.

ⓐ \({8}^{{\text{log}}_{8}7}\) ⓑ \({\text{log}}_{6}{6}^{-2}\)

Try it.

ⓐ \({6}^{{\text{log}}_{6}15}\) ⓑ \({\text{log}}_{8}{8}^{-4}\)

Solution

ⓐ 15 ⓑ \(-4\)

Try it.

ⓐ \({10}^{\text{log}\sqrt{5}}\) ⓑ \(\text{log}{10}^{-2}\)

Try it.

ⓐ \({10}^{\text{log}\sqrt{3}}\) ⓑ \(\text{log}{10}^{-1}\)

Solution

ⓐ \(\sqrt{3}\) ⓑ \(-1\)

Try it.

ⓐ \({e}^{\text{ln}4}\) ⓑ \(\text{ln}\ {e}^{2}\)

Try it.

ⓐ \({e}^{\text{ln}3}\) ⓑ \(\text{ln}\ {e}^{7}\)

Solution

ⓐ 3 ⓑ 7

In the following exercises, use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify if possible.

Try it.

\({\text{log}}_{4}6x\)

Try it.

\({\text{log}}_{5}8y\)

Solution

\({\text{log}}_{5}8+{\text{log}}_{5}y\)

Try it.

\({\text{log}}_{2}32xy\)

Try it.

\({\text{log}}_{3}81xy\)

Solution

\(4+{\text{log}}_{3}x+{\text{log}}_{3}y\)

Try it.

\(\text{log}100x\)

Try it.

\(\text{log}1000y\)

Solution

\(3+\text{log}y\)

In the following exercises, use the Quotient Property of Logarithms to write each logarithm as a sum of logarithms. Simplify if possible.

Try it.

\({\text{log}}_{3}\frac{3}{8}\)

Try it.

\({\text{log}}_{6}\frac{5}{6}\)

Solution

\({\text{log}}_{6}5-1\)

Try it.

\({\text{log}}_{4}\frac{16}{y}\)

Try it.

\({\text{log}}_{5}\frac{125}{x}\)

Solution

\(3-{\text{log}}_{5}x\)

Try it.

\(\text{log}\frac{x}{10}\)

Try it.

\(\text{log}\frac{10,000}{y}\)

Solution

\(4-\text{log}y\)

Try it.

\(\text{ln}\frac{{e}^{3}}{3}\)

Try it.

\(\text{ln}\frac{{e}^{4}}{16}\)

Solution

\(4-\text{ln}16\)

In the following exercises, use the Power Property of Logarithms to expand each. Simplify if possible.

Try it.

\({\text{log}}_{3}{x}^{2}\)

Try it.

\({\text{log}}_{2}{x}^{5}\)

Solution

\(5{\text{log}}_{2}x\)

Try it.

\(\text{log}{x}^{-2}\)

Try it.

\(\text{log}{x}^{-3}\)

Solution

\(-3\text{log}\ x\)

Try it.

\({\text{log}}_{4}\sqrt{x}\)

Try it.

\({\text{log}}_{5}\sqrt[3]{x}\)

Solution

\(\frac{1}{3}{\text{log}}_{5}x\)

Try it.

\(\text{ln}\ {x}^{\sqrt{3}}\)

Try it.

\(\text{ln}\ {x}^{\sqrt[3]{4}}\)

Solution

\(\sqrt[3]{4}\text{ln}\ x\)

In the following exercises, use the Properties of Logarithms to expand the logarithm. Simplify if possible.

Try it.

\({\text{log}}_{5}(4{x}^{6}{y}^{4})\)

Try it.

\({\text{log}}_{2}(3{x}^{5}{y}^{3})\)

Solution

\({\text{log}}_{2}3+5{\text{log}}_{2}x+3{\text{log}}_{2}y\)

Try it.

\({\text{log}}_{3}(\sqrt{2}{x}^{2})\)

Try it.

\({\text{log}}_{5}(\sqrt[4]{21}{y}^{3})\)

Solution

\(\frac{1}{4}{\text{log}}_{5}21+3{\text{log}}_{5}y\)

Try it.

\({\text{log}}_{3}\frac{x{y}^{2}}{{z}^{2}}\)

Try it.

\({\text{log}}_{5}\frac{4a{b}^{3}{c}^{4}}{{d}^{2}}\)

Solution

\({\text{log}}_{5}4+{\text{log}}_{5}a+3{\text{log}}_{5}b\)
\(+\ 4{\text{log}}_{5}c-2{\text{log}}_{5}d\)

Try it.

\({\text{log}}_{4}\frac{\sqrt{x}}{16{y}^{4}}\)

Try it.

\({\text{log}}_{3}\frac{\sqrt[3]{{x}^{2}}}{27{y}^{4}}\)

Solution

\(\frac{2}{3}{\text{log}}_{3}x-3-4{\text{log}}_{3}y\)

Try it.

\({\text{log}}_{2}\frac{\sqrt{2x+{y}^{2}}}{{z}^{2}}\)

Try it.

\({\text{log}}_{3}\frac{\sqrt{3x+2{y}^{2}}}{5{z}^{2}}\)

Solution

\(\frac{1}{2}{\text{log}}_{3}(3x+2{y}^{2})-{\text{log}}_{3}5-2{\text{log}}_{3}z\)

Try it.

\({\text{log}}_{2}\sqrt[4]{\frac{5{x}^{3}}{2{y}^{2}{z}^{4}}}\)

Try it.

\({\text{log}}_{5}\sqrt[3]{\frac{3{x}^{2}}{4{y}^{3}z}}\)

Solution

\(\frac{1}{3}({\text{log}}_{5}3+2{\text{log}}_{5}x-{\text{log}}_{5}4\)
\(-\ 3{\text{log}}_{5}y-{\text{log}}_{5}z)\)

In the following exercises, use the Properties of Logarithms to condense the logarithm. Simplify if possible.

Try it.

\({\text{log}}_{6}4+{\text{log}}_{6}9\)

Try it.

\(\text{log}4+\text{log}25\)

Solution

2

Try it.

\({\text{log}}_{2}80-{\text{log}}_{2}5\)

Try it.

\({\text{log}}_{3}36-{\text{log}}_{3}4\)

Solution

2

Try it.

\({\text{log}}_{3}4+{\text{log}}_{3}(x+1)\)

Try it.

\({\text{log}}_{2}5-{\text{log}}_{2}(x-1)\)

Solution

\({\text{log}}_{2}\frac{5}{x-1}\)

Try it.

\({\text{log}}_{7}3+{\text{log}}_{7}x-{\text{log}}_{7}y\)

Try it.

\({\text{log}}_{5}2-{\text{log}}_{5}x-{\text{log}}_{5}y\)

Solution

\({\text{log}}_{5}\frac{2}{xy}\)

Try it.

\(4{\text{log}}_{2}x+6{\text{log}}_{2}y\)

Try it.

\(6{\text{log}}_{3}x+9{\text{log}}_{3}y\)

Solution

\({\text{log}}_{3}{x}^{6}{y}^{9}\)

Try it.

\({\text{log}}_{3}({x}^{2}-1)-2{\text{log}}_{3}(x-1)\)

Try it.

\(\text{log}({x}^{2}+2x+1)-2\text{log}(x+1)\)

Solution

0

Try it.

\(4\text{log}\ x-2\text{log}y-3\text{log}z\)

Try it.

\(3\text{ln}\ x+4\text{ln}\ y-2\text{ln}\ z\)

Solution

\(\text{ln}\frac{{x}^{3}{y}^{4}}{{z}^{2}}\)

Try it.

\(\frac{1}{3}\text{log}\ x-3\text{log}(x+1)\)

Try it.

\(2\text{log}(2x+3)+\frac{1}{2}\text{log}(x+1)\)

Solution

\(\text{log}{(2x+3)}^{2}\cdot \sqrt{x+1}\)

Use the Change-of-Base Formula

In the following exercises, use the Change-of-Base Formula, rounding to three decimal places, to approximate each logarithm.

Try it.

\({\text{log}}_{3}42\)

Try it.

\({\text{log}}_{5}46\)

Solution

\(2.379\)

Try it.

\({\text{log}}_{12}87\)

Try it.

\({\text{log}}_{15}93\)

Solution

\(1.674\)

Try it.

\({\text{log}}_{\sqrt{2}}17\)

Try it.

\({\text{log}}_{\sqrt{3}}21\)

Solution

\(5.542\)

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. Evaluate: ⓐ \({a}^{0}\) ⓑ \({a}^{1}.\)
   If you missed this problem, review .

   Одкриј го одговорот

   ⓐ \(1\); ⓑ \(a\)

2. Write with a rational exponent: \(\sqrt[3]{{x}^{2}y}.\)
   If you missed this problem, review .

   Одкриј го одговорот

   \({\left({x}^{2}y\right)}^{\frac{1}{3}}\)

3. Round to three decimal places: 2.5646415.
   If you missed this problem, review .

   Одкриј го одговорот

   \(2.565\)

4. Evaluate using the properties of logarithms: ⓐ \({\text{log}}_{8}1\) and ⓑ \({\text{log}}_{6}6.\)

   Одкриј го одговорот

   ⓐ
   

   	\(\ {\text{log}}_{8}1\)
   Use the property, \({\text{log}}_{a}1=0\).	\(\ 0\ {\text{log}}_{8}1=0\)

   ⓑ
   \(\begin{array}{llllll} & & & \ {\text{log}}_{6}6 & & \\ \text{Use the property,}\ {\text{log}}_{a}a=1. & & & \ 1 & & \ {\text{log}}_{6}6=1\end{array}\)

5. Evaluate using the properties of logarithms: ⓐ \({\text{log}}_{13}1\) ⓑ \({\text{log}}_{9}9.\)

   Одкриј го одговорот

   ⓐ 0 ⓑ 1

6. Evaluate using the properties of logarithms: ⓐ \({\text{log}}_{5}1\) ⓑ \({\text{log}}_{7}7.\)

   Одкриј го одговорот

   ⓐ 0 ⓑ 1

7. Evaluate using the properties of logarithms: ⓐ \({4}^{{\text{log}}_{4}9}\) and ⓑ \({\text{log}}_{3}{3}^{5}.\)

   Одкриј го одговорот

   ⓐ
   

   	\(\ {4}^{{\text{log}}_{4}9}\)
   Use the property, \({a}^{{\text{log}}_{a}x}=x\).	\(\ 9\ {4}^{{\text{log}}_{4}9}=9\)

   ⓑ
   

   	\(\ {\text{log}}_{3}{3}^{5}\)
   Use the property, \({a}^{{\text{log}}_{a}x}=x\).	\(\ 5\ {\text{log}}_{3}{3}^{5}=5\)

8. Evaluate using the properties of logarithms: ⓐ \({5}^{{\text{log}}_{5}15}\) ⓑ \({\text{log}}_{7}{7}^{4}.\)

   Одкриј го одговорот

   ⓐ 15 ⓑ 4

9. Evaluate using the properties of logarithms: ⓐ \({2}^{{\text{log}}_{2}8}\) ⓑ \({\text{log}}_{2}{2}^{15}.\)

   Одкриј го одговорот

   ⓐ 8 ⓑ 15

10. Use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify, if possible: ⓐ \({\text{log}}_{3}7x\) and ⓑ \({\text{log}}_{4}64xy.\)

    Одкриј го одговорот

    ⓐ
    

    	\(\ {\text{log}}_{3}7x\)
    Use the Product Property, \({\text{log}}_{a}(M\cdot N)={\text{log}}_{a}M+{\text{log}}_{a}N\).	\(\ {\text{log}}_{3}7+{\text{log}}_{3}x\)
    	\(\ {\text{log}}_{3}7x={\text{log}}_{3}7+{\text{log}}_{3}x\)

    ⓑ
    

    	\(\ {\text{log}}_{4}64xy\)
    Use the Product Property, \({\text{log}}_{a}(M\cdot N)={\text{log}}_{a}M+{\text{log}}_{a}N\).	\(\ {\text{log}}_{4}64+{\text{log}}_{4}x+{\text{log}}_{4}y\)
    Simplify by evaluating \({\text{log}}_{4}64\).	\(\ 3+{\log }_{4}x+{\log }_{4}y\)
    	\(\ {\text{log}}_{4}64xy=3+{\text{log}}_{4}x+{\text{log}}_{4}y\)

11. Use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify, if possible.

    ⓐ \({\text{log}}_{3}3x\) ⓑ \({\text{log}}_{2}8xy\)

    Одкриј го одговорот

    ⓐ \(1+{\text{log}}_{3}x\)
    ⓑ \(3+{\text{log}}_{2}x+{\text{log}}_{2}y\)

12. Use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify, if possible.

    ⓐ \({\text{log}}_{9}9x\) ⓑ \({\text{log}}_{3}27xy\)

    Одкриј го одговорот

    ⓐ \(1+{\text{log}}_{9}x\)
    ⓑ \(3+{\text{log}}_{3}x+{\text{log}}_{3}y\)

13. Use the Quotient Property of Logarithms to write each logarithm as a difference of logarithms. Simplify, if possible.
    ⓐ \({\text{log}}_{5}\frac{5}{7}\) and ⓑ \(\text{log}\frac{x}{100}\)

    Одкриј го одговорот

    ⓐ
    

    	\(\ {\text{log}}_{5}\frac{5}{7}\)
    Use the Quotient Property, \({\text{log}}_{a}\frac{M}{N}={\text{log}}_{a}M-{\text{log}}_{a}N\).	\(\ {\text{log}}_{5}5-{\text{log}}_{5}7\)
    Simplify.	\(\ 1-{\text{log}}_{5}7\)
    	\(\ {\text{log}}_{5}\frac{5}{7}=1-{\text{log}}_{5}7\)

    ⓑ
    

    	\(\ \text{log}\frac{x}{100}\)
    Use the Quotient Property, \({\text{log}}_{a}\frac{M}{N}={\text{log}}_{a}M-{\text{log}}_{a}N\).	\(\ \text{log}\ x-\text{log}100\)
    Simplify.	\(\ \text{log}\ x-2\)
    	\(\ \text{log}\frac{x}{100}=\text{log}\ x-2\)

14. Use the Quotient Property of Logarithms to write each logarithm as a difference of logarithms. Simplify, if possible.

    ⓐ \({\text{log}}_{4}\frac{3}{4}\) ⓑ \(\text{log}\frac{x}{1000}\)

    Одкриј го одговорот

    ⓐ \({\text{log}}_{4}3-1\) ⓑ \(\text{log}\ x-3\)

15. Use the Quotient Property of Logarithms to write each logarithm as a difference of logarithms. Simplify, if possible.

    ⓐ \({\text{log}}_{2}\frac{5}{4}\) ⓑ \(\text{log}\frac{10}{y}\)

    Одкриј го одговорот

    ⓐ \({\text{log}}_{2}5-2\) ⓑ \(1-\text{log}y\)

16. Use the Power Property of Logarithms to write each logarithm as a product of logarithms. Simplify, if possible.
    ⓐ \({\text{log}}_{5}{4}^{3}\) and ⓑ \(\text{log}{x}^{10}\)

    Одкриј го одговорот

    ⓐ
    

    	\(\ {\text{log}}_{5}{4}^{3}\)
    Use the Power Property, \({\text{log}}_{a}{M}^{p}=p\ {\text{log}}_{a}M\).	\(\ 3{\text{log}}_{5}4\)
    	\(\ {\text{log}}_{5}{4}^{3}=3{\text{log}}_{5}4\)

    ⓑ
    

    	\(\ \text{log}{x}^{10}\)
    Use the Power Property, \({\text{log}}_{a}{M}^{p}=p\ {\text{log}}_{a}M\).	\(\ 10\ \text{log}\ x\)
    	\(\ \text{log}{x}^{10}=10\text{log}\ x\)

17. Use the Power Property of Logarithms to write each logarithm as a product of logarithms. Simplify, if possible.

    ⓐ \({\text{log}}_{7}{5}^{4}\) ⓑ \(\text{log}{x}^{100}\)

    Одкриј го одговорот

    ⓐ \(4{\text{log}}_{7}5\) ⓑ \(100\cdot \text{log}\ x\)

18. Use the Power Property of Logarithms to write each logarithm as a product of logarithms. Simplify, if possible.

    ⓐ \({\text{log}}_{2}{3}^{7}\) ⓑ \(\text{log}{x}^{20}\)

    Одкриј го одговорот

    ⓐ \(7{\text{log}}_{2}3\) ⓑ \(20\cdot \text{log}\ x\)

19. Use the Properties of Logarithms to expand the logarithm \({\text{log}}_{4}(2{x}^{3}{y}^{2})\). Simplify, if possible.

    Одкриј го одговорот

    	\(\ {\text{log}}_{4}(2{x}^{3}{y}^{2})\)
    Use the Product Property, \({\text{log}}_{a}M\cdot N={\text{log}}_{a}M+{\text{log}}_{a}N\).	\(\ {\text{log}}_{4}2+{\text{log}}_{4}{x}^{3}+{\text{log}}_{4}{y}^{2}\)
    Use the Power Property, \({\text{log}}_{a}{M}^{p}=p\ {\text{log}}_{a}M\), on the last two terms.	\(\ {\text{log}}_{4}2+3{\text{log}}_{4}x+2{\text{log}}_{4}y\)
    Simplify.	\(\ \frac{1}{2}+3{\text{log}}_{4}x+2{\text{log}}_{4}y\)
    	\(\ {\text{log}}_{4}(2{x}^{3}{y}^{2})=\frac{1}{2}+3{\text{log}}_{4}x+2{\text{log}}_{4}y\)

20. Use the Properties of Logarithms to expand the logarithm \({\text{log}}_{2}(5{x}^{4}{y}^{2})\). Simplify, if possible.

    Одкриј го одговорот

    \({\text{log}}_{2}5+4{\text{log}}_{2}x+2{\text{log}}_{2}y\)

21. Use the Properties of Logarithms to expand the logarithm \({\text{log}}_{3}(7{x}^{5}{y}^{3})\). Simplify, if possible.

    Одкриј го одговорот

    \({\text{log}}_{3}7+5{\text{log}}_{3}x+3{\text{log}}_{3}y\)

22. Use the Properties of Logarithms to expand the logarithm \({\text{log}}_{2}\sqrt[4]{\frac{{x}^{3}}{3{y}^{2}z}}\). Simplify, if possible.

    Одкриј го одговорот

    	\(\ {\text{log}}_{2}\sqrt[4]{\frac{{x}^{3}}{3{y}^{2}z}}\)
    Rewrite the radical with a rational exponent.	\(\ {\text{log}}_{2}{(\frac{{x}^{3}}{3{y}^{2}z})}^{\frac{1}{4}}\)
    Use the Power Property, \({\text{log}}_{a}{M}^{p}=p\ {\text{log}}_{a}M\).	\(\ \frac{1}{4}{\text{log}}_{2}(\frac{{x}^{3}}{3{y}^{2}z})\)
    Use the Quotient Property, \({\text{log}}_{a}M\cdot N={\text{log}}_{a}M-{\text{log}}_{a}N\).	\(\ \frac{1}{4}({\text{log}}_{2}({x}^{3})-{\text{log}}_{2}(3{y}^{2}z))\)
    Use the Product Property, \({\text{log}}_{a}M\cdot N={\text{log}}_{a}M+{\text{log}}_{a}N\), in the second term.	\(\ \frac{1}{4}({\text{log}}_{2}({x}^{3})-({\text{log}}_{2}3+{\text{log}}_{2}{y}^{2}+{\text{log}}_{2}z))\)
    Use the Power Property, \({\text{log}}_{a}{M}^{p}=p\ {\text{log}}_{a}M\), inside the parentheses.	\(\ \frac{1}{4}(3{\text{log}}_{2}x-({\text{log}}_{2}3+2{\text{log}}_{2}y+{\text{log}}_{2}z))\)
    Simplify by distributing.	\(\ \frac{1}{4}(3{\text{log}}_{2}x-{\text{log}}_{2}3-2{\text{log}}_{2}y-{\text{log}}_{2}z)\)
    	\({\text{log}}_{2}\sqrt[4]{\frac{{x}^{3}}{3{y}^{2}z}}=\frac{1}{4}(3{\text{log}}_{2}x-{\text{log}}_{2}3-2{\text{log}}_{2}y-{\text{log}}_{2}z)\)

23. Use the Properties of Logarithms to expand the logarithm \({\text{log}}_{4}\sqrt[5]{\frac{{x}^{4}}{2{y}^{3}{z}^{2}}}\). Simplify, if possible.

    Одкриј го одговорот

    \(\frac{1}{5}(4{\text{log}}_{4}x-\frac{1}{2}-3{\text{log}}_{4}y-2{\text{log}}_{4}z)\)

24. Use the Properties of Logarithms to expand the logarithm \({\text{log}}_{3}\sqrt[3]{\frac{{x}^{2}}{5{y}^{}z}}\). Simplify, if possible.

    Одкриј го одговорот

    \(\frac{1}{3}(2{\text{log}}_{3}x-{\text{log}}_{3}5-{\text{log}}_{3}y-{\text{log}}_{3}z)\)

25. Use the Properties of Logarithms to condense the logarithm \({\text{log}}_{4}3+{\text{log}}_{4}x-{\text{log}}_{4}y\). Simplify, if possible.

    Одкриј го одговорот

    The log expressions all have the same base, 4.	\(\ {\text{log}}_{4}3+{\text{log}}_{4}x-{\text{log}}_{4}y\)
    The first two terms are added, so we use the Product Property, \({\text{log}}_{a}M+{\text{log}}_{a}N={\text{log}}_{a}M\cdot N\).	\(\ {\text{log}}_{4}3x-{\text{log}}_{4}y\)
    Since the logs are subtracted, we use the Quotient Property, \({\text{log}}_{a}M-{\text{log}}_{a}N={\text{log}}_{a}\frac{M}{N}\).	\(\ {\text{log}}_{4}\frac{3x}{y}\)
    	\(\ {\text{log}}_{4}3+{\text{log}}_{4}x-{\text{log}}_{4}y={\text{log}}_{4}\frac{3x}{y}\)

26. Use the Properties of Logarithms to condense the logarithm \({\text{log}}_{2}5+{\text{log}}_{2}x-{\text{log}}_{2}y\). Simplify, if possible.

    Одкриј го одговорот

    \({\text{log}}_{2}\frac{5x}{y}\)

27. Use the Properties of Logarithms to condense the logarithm \({\text{log}}_{3}6-{\text{log}}_{3}x-{\text{log}}_{3}y\). Simplify, if possible.

    Одкриј го одговорот

    \({\text{log}}_{3}\frac{6}{xy}\)

28. Use the Properties of Logarithms to condense the logarithm \(2{\text{log}}_{3}x+4{\text{log}}_{3}(x+1)\). Simplify, if possible.

    Одкриј го одговорот

    The log expressions have the same base, 3.	\(\ 2{\text{log}}_{3}x+4{\text{log}}_{3}(x+1)\)
    Use the Power Property, \({\text{log}}_{a}M+{\text{log}}_{a}N={\text{log}}_{a}M\cdot N\).	\(\ {\text{log}}_{3}{x}^{2}+{\text{log}}_{3}{(x+1)}^{4}\)
    The terms are added, so we use the Product Property, \({\text{log}}_{a}M+{\text{log}}_{a}N={\text{log}}_{a}M\cdot N\).	\(\ {\text{log}}_{3}{x}^{2}{(x+1)}^{4}\)
    	\(\ 2{\text{log}}_{3}x+4{\text{log}}_{3}(x+1)={\text{log}}_{3}{x}^{2}{(x+1)}^{4}\)

29. Use the Properties of Logarithms to condense the logarithm \(3{\text{log}}_{2}x+2{\text{log}}_{2}(x-1)\). Simplify, if possible.

    Одкриј го одговорот

    \({\text{log}}_{2}{x}^{3}{(x-1)}^{2}\)

30. Use the Properties of Logarithms to condense the logarithm \(2\text{log}\ x+2\text{log}(x+1)\). Simplify, if possible.

    Одкриј го одговорот

    \(\text{log}{x}^{2}{(x+1)}^{2}\)

31. Rounding to three decimal places, approximate \({\text{log}}_{4}35.\)

    Одкриј го одговорот

    Use the Change-of-Base Formula.
    Identify a and M. Choose 10 for b.
    Enter the expression \(\frac{\text{log}35}{\text{log}4}\) in the calculator using the log button for base 10. Round to three decimal places.

32. Rounding to three decimal places, approximate \({\text{log}}_{3}42.\)

    Одкриј го одговорот

    \(3.402\)

33. Rounding to three decimal places, approximate \({\text{log}}_{5}46.\)

    Одкриј го одговорот

    \(2.379\)

34. ⓐ \({\text{log}}_{4}1\) ⓑ \({\text{log}}_{8}8\)

35. ⓐ \({\text{log}}_{12}1\) ⓑ \(\text{ln}\ e\)

    Одкриј го одговорот

    ⓐ 0 ⓑ 1

36. ⓐ \({3}^{{\text{log}}_{3}6}\) ⓑ \({\text{log}}_{2}{2}^{7}\)

37. ⓐ \({5}^{{\text{log}}_{5}10}\) ⓑ \({\text{log}}_{4}{4}^{10}\)

    Одкриј го одговорот

    ⓐ 10 ⓑ 10

38. ⓐ \({8}^{{\text{log}}_{8}7}\) ⓑ \({\text{log}}_{6}{6}^{-2}\)

39. ⓐ \({6}^{{\text{log}}_{6}15}\) ⓑ \({\text{log}}_{8}{8}^{-4}\)

    Одкриј го одговорот

    ⓐ 15 ⓑ \(-4\)

40. ⓐ \({10}^{\text{log}\sqrt{5}}\) ⓑ \(\text{log}{10}^{-2}\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\approx

approximately equal
Equal to the precision shown, not exactly.

\neq

not equal
The two sides are different.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

i

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Use the Properties of Logarithms

1. Use the properties of logarithms
2. Use the Change of Base Formula
3. For
4. If
5. If
6. If

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

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x²√÷×π( )=∫d/dxlim[ ]sinlogeˣ

Parts of this page are adapted from OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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