## 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 $x^3$.

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 ${x^2+1 \over \log_{10} 5}.$