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].
We have already seen that the logarithm of a product is the sum of the logarithms of the factors:
[latex]\displaystyle
\log_b(xy) = \log_bx + \log_by[/latex]
Similarly, the logarithm of the ratio of two quantities is the difference of the logarithms:
[latex]\displaystyle
\log_b\left( \frac{x}{y}\right) = \log_bx -
log_by.[/latex]
We can show that this is true by the following example:
Let [latex]u=\log_b x[/latex] and [latex]v=\log_b y[/latex].
Then [latex]b^u=x[/latex] and [latex]b^v=y.[/latex]
Then:
[latex]\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}[/latex]
Another way to show that this rule is true, is to apply both the power and product rules and the fact that dividing by [latex]y[/latex] is the same is multiplying by [latex]y^{-1}.[/latex] So we can write:
[latex]\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}[/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.
Most common scientific calculators have a key for computing logarithms with base [latex]10[/latex], but do not have keys for other bases. So, if you needed to get an approximation to a number like [latex]\log_4(9)[/latex] it can be difficult to do so. One could easily guess that it is between [latex]1[/latex] and [latex]2[/latex] since [latex]9[/latex] is between [latex]4^1[/latex] and [latex]4^2[/latex], 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:
[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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