Working With Logarithms

Logarithms of Products

A useful property of logarithms states that the logarithm of a product of two quantities is the sum of the logarithms of the two factors. In symbols,

\log_b(xy)=\log_b(x)+\log_b(y).

Learning Objectives

Relate the product rule for logarithms to the rules for operating with exponents, and use this rule to rewrite logarithms of products

Key Takeaways

Key Points

The logarithm of a product is the sum of the logarithms of the factors.
The product rule does not apply when the base of the two logarithms are different.

Key Terms

exponent: The power to which a number, symbol, or expression is to be raised. For example, the 3 in $x^3$ .

Logarithms

The logarithm of a number is the exponent by which another fixed value, the base, has to be raised to produce that number. For example, the logarithm of

1000

in base

10

3

, because

10^3=1000.

More generally, if

x=b^y

, then

y

is the logarithm base

b

x

, written:

y=\log_b(x)

, so

\log_{10}(1000)=3

. It is useful to think of logarithms as inverses of exponentials. So, for example:

\displaystyle \log_b(b^z)=z

And:

\displaystyle b^{\log_b(z)}=z

Product Rule for Logarithms

Logarithms were introduced by John Napier in the early 17th century as a means to simplify calculations. Logarithms were rapidly adopted by navigators, scientists, engineers, and others to perform computations more easily by using slide rules and logarithm tables. Tedious multi-digit multiplication steps can be replaced by table look-ups and simpler addition, because of the fact that the logarithm of a product is the sum of the logarithms of the factors:

\displaystyle log_b(xy) = log_b(x) + log_b(y)

We can see that this rule is true by writing the logarithms in terms of exponentials. Let

\log_b(x)=v

and

\log_b(y)=w.

Writing these equations as exponentials:

\displaystyle b^v=x

And:

\displaystyle b^w=y.

Then note that:

\displaystyle \begin{align} xy&=b^vb^w\\ &=b^{v+w} \end{align}

Taking the logarithm base

b

of both sides of this last equation yields:

\displaystyle \begin{align} \log_b(xy)&=\log_b(b^{v+w})\\ &=v+w\\ &=\log_b(x) + \log_b(y) \end{align}

This is a very useful property of logarithms, because it can sometimes simplify more complex expressions. For example:

\displaystyle \log_{10}(10^x\cdot 100^{x^3+1})=\log_{10} (10^x)+\log_{10}(100^{x^3+1})

Then because

100

10^2

, we have:

\displaystyle \begin{align} x+\log_{10}(10^{2(x^3+1)}) &= x+2(x^3+1)\\ &=2x^3+x+2 \end{align}

Logarithms of Powers

The logarithm of the

p\text{th}

power of a quantity is

p

times the logarithm of the quantity. In symbols,

\log_b(x^p)=p\log_b(x).

Learning Objectives

Relate the power rule for logarithms to the rules for operating with exponents, and use this rule to rewrite logarithms of powers

Key Takeaways

Key Points

The logarithm of a product is the sum of the logarithms of the factors.
An exponent, $p$ , signifies that a number is being multiplied by itself $p$ number of times. Because the logarithm of a product is the sum of the logarithms of the factors, the logarithm of a number, $x$ , to an exponent, $p$ , is the same as the logarithm of $x$ added together $p$ times, so it is equal to $p\log_b(x).$

Key Terms

base: A number raised to the power of an exponent.
logarithm: The logarithm of a number is the exponent by which another fixed value, the base, has to be raised to produce that number.
exponent: The power to which a number, symbol, or expression is to be raised. For example, the 3 in $x^3$ .

The Power Rule for Logarithms

We have already seen that the logarithm of a product is the sum of the logarithms of the factors:

\displaystyle \log _b \left( {xy} \right) = \log _b \left( x \right) + \log _b \left( y \right)

If we apply this rule repeatedly we can devise another rule for simplifying expressions of the form

\log_b x^p

. Recall that

x^p

can be thought of as

x \cdot x \cdot x \cdots x

where there are

p

factors of

x

. Then we have:

\displaystyle \begin{align} \log_b(x^p) &= \log_b (x \cdot x \cdots x) \\ &= \log_b x + \log_b x + \cdots +\log_b x \\ &= p\log_b x \end{align}

Since the

p

factors of

x

are converted to

p

summands by the product rule formula.

Example 1: Simplify the expression $\log_3(3^x\cdot 9x^{100})$

First expand the log:

\displaystyle \log_3(3^x\cdot 9x^{100}) =\log_3 (3^x) + \log_3 9 + \log_3(x^{100})

Next use the product and power rule to simplify:

\displaystyle \log_3 (3^x) + \log_3 9 + \log_3 (x^{100})= x+2+100\log_3 x

Example 2: Solve $2^{(x+1)}=10^3$ for $x$ using logarithms

Start by taking the logarithm with base

2

of both sides:

\displaystyle \begin{align} \log_2 (2^{(x+1)}) &= \log_2 (10^3)\\ x+1&=3\log_2(10)\\ x&=3\log_2(10)-1 \end{align}

Therefore a solution would be

x=3\log_2(10) -1.

Logarithms of Quotients

The logarithm of the ratio of two quantities is the difference of the logarithms of the quantities. In symbols,

\log_b\left( \frac{x}{y}\right) = \log_bx - \log_by.

Learning Objectives

Relate the quotient rule for logarithms to the rules for operating with exponents, and use this rule to rewrite logarithms of quotients

Key Takeaways

Key Points

The logarithm of a number is the exponent by which another fixed value, the base, has to be raised to produce that number.
The logarithm of a product is the sum of the logarithms of the factors.
The logarithm of the ratio or quotient of two numbers is the difference of the logarithms.

Key Terms

exponent: The power to which a number, symbol, or expression is to be raised. For example, the 3 in $x^3$ .

We have already seen that the logarithm of a product is the sum of the logarithms of the factors:

\displaystyle \log_b(xy) = \log_bx + \log_by

Similarly, the logarithm of the ratio of two quantities is the difference of the logarithms:

\displaystyle \log_b\left( \frac{x}{y}\right) = \log_bx - log_by.

We can show that this is true by the following example: Let

u=\log_b x

and

v=\log_b y

. Then

b^u=x

and

b^v=y.

Then:

\displaystyle \begin{align} \log_b\left(\frac{x}{y}\right)&=\log_b\left({b^u \over b^v}\right)\\ &= \log_b(b^{u-v}) \\ &=u-v\\ &= \log_b x - \log_b y \end{align}

Another way to show that this rule is true, is to apply both the power and product rules and the fact that dividing by

y

is the same is multiplying by

y^{-1}.

So we can write:

\displaystyle \begin{align} \log_b\left(\frac{x}{y}\right)&=\log_b(x\cdot y^{-1})\\ & = \log_bx + \log_b(y^{-1})\\& = \log_bx -\log_by \end{align}

Example: write the expression $\log_2\left({x^4y^9 \over z^{100}}\right)$ in a simpler way

By applying the product, power, and quotient rules, you could write this expression as:

\displaystyle \log_2(x^4)+\log_2(y^9)-\log_2(z^{100}) = 4\log_2x+9\log_2y-100\log_2z.

Changing Logarithmic Bases

A logarithm written in one base can be converted to an equal quantity written in a different base.

Learning Objectives

Use the change of base formula to convert logarithms to different bases

Key Takeaways

Key Points

The base of a logarithm can be changed by expressing it as the quotient of two logarithms with a common base.
Changing a logarithm's base to $10$ makes it much simpler to evaluate; it can be done on a calculator.

Key Terms

logarithm: The logarithm of a number is the exponent by which another fixed value, the base, has to be raised to produce that number.
base: A number raised to the power of an exponent.

Most common scientific calculators have a key for computing logarithms with base

10

, but do not have keys for other bases. So, if you needed to get an approximation to a number like

\log_4(9)

it can be difficult to do so. One could easily guess that it is between

1

and

2

since

9

is between

4^1

and

4^2

, but it is difficult to get an accurate approximation. Fortunately, there is a change of base formula that can help.

Change of Base Formula

The change of base formula for logarithms is:

\displaystyle \log_a(x)=\frac{\log_b(x)}{\log_b(a)}

Thus, for example, we could calculate that

\log_4(9)=\frac{\log_{10}(9)}{\log_{10}(4)}

which could be computed on almost any handheld calculator.

Deriving the Formula

To see why the formula is true, give

\log_a(x)

a name like

z

\displaystyle z=\log_a(x)

Write this as

a^z=x

Now take the logarithm with base

b

of both sides, yielding:

\displaystyle \log_b a^z = \log_bx

Using the power rule gives:

\displaystyle z \cdot \log_ba = \log_b x

Dividing both sides by

\log_ba

gives:

\displaystyle z={\log_b x \over \log_ba}.

Thus we have

\log_a x ={\log_b x \over \log_b a}.

Example

An expression of the form

\log_5(10^{x^2+1})

might be easier to graph on a graphing calculator or other device if it were written in base

10

instead of base 5. The change-of-base formula can be applied to it:

\displaystyle \log_5(10^{x^2+1}) = {\log_{10}(10^{x^2+1}) \over \log_{10}5}

Which can be written as

{x^2+1 \over \log_{10} 5}.

Working With Logarithms

Logarithms of Products

Learning Objectives

Key Takeaways

Key Points

Key Terms

Logarithms

Product Rule for Logarithms

Logarithms of Powers

Learning Objectives

Key Takeaways

Key Points

Key Terms

The Power Rule for Logarithms

Example 1: Simplify the expression $\log_3(3^x\cdot 9x^{100})$

Example 2: Solve $2^{(x+1)}=10^3$ for $x$ using logarithms

Logarithms of Quotients

Learning Objectives

Key Takeaways

Key Points

Key Terms

Example: write the expression $\log_2\left({x^4y^9 \over z^{100}}\right)$ in a simpler way

Changing Logarithmic Bases

Learning Objectives

Key Takeaways

Key Points

Key Terms

Change of Base Formula

Deriving the Formula

Example

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Study Guides > Boundless Algebra

Working With Logarithms

Logarithms of Products

Learning Objectives

Key Takeaways

Key Points

Key Terms

Logarithms

Product Rule for Logarithms

Logarithms of Powers

Learning Objectives

Key Takeaways

Key Points

Key Terms

The Power Rule for Logarithms

Example 1: Simplify the expression log⁡3(3x⋅9x100)\log_3(3^x\cdot 9x^{100})log3​(3x⋅9x100)

Example 2: Solve 2(x+1)=1032^{(x+1)}=10^32(x+1)=103 for xxx using logarithms

Logarithms of Quotients

Learning Objectives

Key Takeaways

Key Points

Key Terms

Example: write the expression log⁡2(x4y9z100)\log_2\left({x^4y^9 \over z^{100}}\right)log2​(z100x4y9​) in a simpler way

Changing Logarithmic Bases

Learning Objectives

Key Takeaways

Key Points

Key Terms

Change of Base Formula

Deriving the Formula

Example

Licenses & Attributions

CC licensed content, Shared previously

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Example 1: Simplify the expression $\log_3(3^x\cdot 9x^{100})$

Example 2: Solve $2^{(x+1)}=10^3$ for $x$ using logarithms

Example: write the expression $\log_2\left({x^4y^9 \over z^{100}}\right)$ in a simpler way