# 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 $10$

is $3$

, because $10^3=1000.$

More generally, if

$x=b^y$

, then $y$

is the logarithm base $b$

of $x$

, 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

$100$

is $10^2$

, 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$x^3$.

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

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.

$10$

, 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

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

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