# 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, [latex]\log_b(xy)=\log_b(x)+\log_b(y).[/latex]### 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 [latex]x^3[/latex].

### 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 [latex]1000[/latex] in base [latex]10[/latex] is [latex]3[/latex], because [latex]10^3=1000.[/latex] More generally, if [latex]x=b^y[/latex], then [latex]y[/latex] is the logarithm base [latex]b[/latex] of [latex]x[/latex], written: [latex]y=\log_b(x)[/latex], so [latex]\log_{10}(1000)=3[/latex]. It is useful to think of logarithms as inverses of exponentials. So, for example: [latex]\displaystyle \log_b(b^z)=z[/latex] And: [latex]\displaystyle b^{\log_b(z)}=z[/latex]### 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: [latex]\displaystyle log_b(xy) = log_b(x) + log_b(y)[/latex] We can see that this rule is true by writing the logarithms in terms of exponentials. Let [latex]\log_b(x)=v[/latex] and [latex]\log_b(y)=w.[/latex] Writing these equations as exponentials: [latex]\displaystyle b^v=x[/latex] And: [latex]\displaystyle b^w=y.[/latex] Then note that: [latex]\displaystyle \begin{align} xy&=b^vb^w\\ &=b^{v+w} \end{align}[/latex] Taking the logarithm base [latex]b[/latex] of both sides of this last equation yields: [latex]\displaystyle \begin{align} \log_b(xy)&=\log_b(b^{v+w})\\ &=v+w\\ &=\log_b(x) + \log_b(y) \end{align}[/latex] This is a very useful property of logarithms, because it can sometimes simplify more complex expressions. For example: [latex]\displaystyle \log_{10}(10^x\cdot 100^{x^3+1})=\log_{10} (10^x)+\log_{10}(100^{x^3+1}) [/latex] Then because [latex]100[/latex] is [latex]10^2[/latex], we have: [latex]\displaystyle \begin{align} x+\log_{10}(10^{2(x^3+1)}) &= x+2(x^3+1)\\ &=2x^3+x+2 \end{align}[/latex]## Logarithms of Powers

The logarithm of the [latex]p\text{th}[/latex] power of a quantity is [latex]p[/latex] times the logarithm of the quantity. In symbols, [latex]\log_b(x^p)=p\log_b(x).[/latex]### 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, [latex]p[/latex], signifies that a number is being multiplied by itself [latex]p[/latex] number of times. Because the logarithm of a product is the sum of the logarithms of the factors, the logarithm of a number, [latex]x[/latex], to an exponent, [latex]p[/latex], is the same as the logarithm of [latex]x[/latex] added together [latex]p[/latex] times, so it is equal to [latex]p\log_b(x).[/latex]

#### 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 [latex]x^3[/latex].

### The Power Rule for Logarithms

We have already seen that the logarithm of a product is the sum of the logarithms of the factors: [latex]\displaystyle \log _b \left( {xy} \right) = \log _b \left( x \right) + \log _b \left( y \right)[/latex] If we apply this rule repeatedly we can devise another rule for simplifying expressions of the form [latex]\log_b x^p[/latex]. Recall that [latex]x^p[/latex] can be thought of as [latex]x \cdot x \cdot x \cdots x[/latex] where there are [latex]p[/latex] factors of [latex]x[/latex]. Then we have: [latex]\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}[/latex] Since the [latex]p[/latex] factors of [latex]x[/latex] are converted to [latex]p [/latex] summands by the product rule formula.### Example 1: Simplify the expression [latex]\log_3(3^x\cdot 9x^{100})[/latex]

First expand the log: [latex]\displaystyle \log_3(3^x\cdot 9x^{100}) =\log_3 (3^x) + \log_3 9 + \log_3(x^{100}) [/latex] Next use the product and power rule to simplify: [latex]\displaystyle \log_3 (3^x) + \log_3 9 + \log_3 (x^{100})= x+2+100\log_3 x[/latex]### Example 2: Solve [latex]2^{(x+1)}=10^3[/latex] for [latex]x[/latex] using logarithms

Start by taking the logarithm with base [latex]2[/latex] of both sides: [latex]\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}[/latex] Therefore a solution would be [latex]x=3\log_2(10) -1. [/latex]## Logarithms of Quotients

The logarithm of the ratio of two quantities is the difference of the logarithms of the quantities. In symbols, [latex]\log_b\left( \frac{x}{y}\right) = \log_bx - \log_by.[/latex]### 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 [latex]x^3[/latex].

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

By applying the product, power, and quotient rules, you could write this expression as: [latex]\displaystyle \log_2(x^4)+\log_2(y^9)-\log_2(z^{100}) = 4\log_2x+9\log_2y-100\log_2z.[/latex]## 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 [latex]10[/latex] 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.

### Change of Base Formula

The change of base formula for logarithms is: [latex]\displaystyle \log_a(x)=\frac{\log_b(x)}{\log_b(a)}[/latex] Thus, for example, we could calculate that [latex]\log_4(9)=\frac{\log_{10}(9)}{\log_{10}(4)}[/latex] which could be computed on almost any handheld calculator.### Deriving the Formula

To see why the formula is true, give [latex]\log_a(x)[/latex] a name like [latex]z[/latex]: [latex]\displaystyle z=\log_a(x)[/latex] Write this as [latex]a^z=x[/latex] Now take the logarithm with base [latex]b[/latex] of both sides, yielding: [latex]\displaystyle \log_b a^z = \log_bx[/latex] Using the power rule gives: [latex]\displaystyle z \cdot \log_ba = \log_b x[/latex] Dividing both sides by [latex]\log_ba[/latex] gives: [latex]\displaystyle z={\log_b x \over \log_ba}.[/latex] Thus we have [latex]\log_a x ={\log_b x \over \log_b a}. [/latex]### Example

An expression of the form [latex]\log_5(10^{x^2+1})[/latex] might be easier to graph on a graphing calculator or other device if it were written in base [latex]10[/latex] instead of base 5. The change-of-base formula can be applied to it: [latex]\displaystyle \log_5(10^{x^2+1}) = {\log_{10}(10^{x^2+1}) \over \log_{10}5}[/latex] Which can be written as [latex]{x^2+1 \over \log_{10} 5}. [/latex]## Licenses & Attributions

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