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 \(\text{(}\approx \text{)}\).

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.

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まず紙で試してみてください。チェックするには答えを明らかにしてください。確認した答えは各ステップの解法で開きます。

1. Evaluate: ⓐ \({a}^{0}\) ⓑ \({a}^{1}.\)

   それを通して ヒント

   答えを明らかにしろ

   ⓐ \(1\); ⓑ \(a\)

2. Write with a rational exponent: \(\sqrt[3]{{x}^{2}y}.\)

   それを通して ヒント

   答えを明らかにしろ

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

3. Round to three decimal places: 2.5646415.

   それを通して ヒント

   答えを明らかにしろ

   \(2.565\)

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

   それを通して ヒント

   答えを明らかにしろ

   ⓐ
   

   	\(\ {\text{log}}_{8}1\)
   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: ⓐ \({\text{log}}_{13}1\) ⓑ \({\text{log}}_{9}9.\)

   それを通して ヒント

   答えを明らかにしろ

   ⓐ 0 ⓑ 1

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

   それを通して ヒント

   答えを明らかにしろ

   ⓐ 0 ⓑ 1

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

   それを通して ヒント

   答えを明らかにしろ

   ⓐ
   

   	\(\ {4}^{{\text{log}}_{4}9}\)
   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: ⓐ \({5}^{{\text{log}}_{5}15}\) ⓑ \({\text{log}}_{7}{7}^{4}.\)

   それを通して ヒント

   答えを明らかにしろ

   ⓐ 15 ⓑ 4

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

   それを通して ヒント

   答えを明らかにしろ

   ⓐ 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 ⓑ \({\text{log}}_{4}64xy.\)

    それを通して ヒント

    答えを明らかにしろ

    ⓐ
    

    	\(\ {\text{log}}_{3}7x\)
    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.

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

    それを通して ヒント

    答えを明らかにしろ

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

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

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

    それを通して ヒント

    答えを明らかにしろ

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

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 ⓑ \(\text{log}\frac{x}{100}\)

    それを通して ヒント

    答えを明らかにしろ

    ⓐ
    

    	\(\ {\text{log}}_{5}\frac{5}{7}\)
    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.

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

    それを通して ヒント

    答えを明らかにしろ

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

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

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

    それを通して ヒント

    答えを明らかにしろ

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

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 ⓑ \(\text{log}{x}^{10}\)

    それを通して ヒント

    答えを明らかにしろ

    ⓐ
    

    	\(\ {\text{log}}_{5}{4}^{3}\)
    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.

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

    それを通して ヒント

    答えを明らかにしろ

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

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

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

    それを通して ヒント

    答えを明らかにしろ

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

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

    それを通して ヒント

    答えを明らかにしろ

    	\(\ {\text{log}}_{4}(2{x}^{3}{y}^{2})\)
    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.

    それを通して ヒント

    答えを明らかにしろ

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

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

    それを通して ヒント

    答えを明らかにしろ

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

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.

    それを通して ヒント

    答えを明らかにしろ

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

    それを通して ヒント

    答えを明らかにしろ

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

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

    それを通して ヒント

    答えを明らかにしろ

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

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.

    それを通して ヒント

    答えを明らかにしろ

    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.

    それを通して ヒント

    答えを明らかにしろ

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

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.

    それを通して ヒント

    答えを明らかにしろ

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

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

    それを通して ヒント

    答えを明らかにしろ

    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.

    それを通して ヒント

    答えを明らかにしろ

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

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

    それを通して ヒント

    答えを明らかにしろ

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

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

    それを通して ヒント

    答えを明らかにしろ

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

    それを通して ヒント

    答えを明らかにしろ

    \(3.402\)

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

    それを通して ヒント

    答えを明らかにしろ

    \(2.379\)

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

    それを通して ヒント

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

    それを通して ヒント

    答えを明らかにしろ

    ⓐ 0 ⓑ 1

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

    それを通して ヒント

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

    それを通して ヒント

    答えを明らかにしろ

    ⓐ 10 ⓑ 10

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

    それを通して ヒント

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

    それを通して ヒント

    答えを明らかにしろ

    ⓐ 15 ⓑ \(-4\)

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

    それを通して ヒント

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ここで使用された記号

どの記号をタップしても、定義、画像、それぞれの文字の意味が表示されます。

\sqrt{x}square root, n-th root
The non-negative number whose square (n-th power) is x.\approxapproximately equal
Equal to the precision shown, not exactly.\neqnot equal
The two sides are different.|x|absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.\binom{n}{k}binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).\pipi
Ratio of a circle's circumference to its diameter, 3.14159…eEuler's number
2.71828…, the base whose exponential is its own derivative.iimaginary unit
i² = −1.\thetatheta
The usual name for an angle.\log_b xlogarithm, natural log
The exponent b must be raised to for x; ln uses base e.\mathbf{v}vector
A quantity with magnitude and direction; a column of numbers.

どうやって: 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

質問

What is a function, really?

A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.

Why do we need complex numbers?

Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.

このページの一部は、 OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). ここで簡略化して再説明する 誤りは我々の責任だ

ここに Precalculus

Complex numbersPolynomial functionsRational functionsSequences and seriesThe binomial theoremConic sectionsVectorsExponential and logarithmic functionsPolynomial division and the remainder theoremParametric equations and polar coordinates