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, 
logb(xy)=logb(x)+logb(y).\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
    x3x^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
10001000
in base
1010
is
33
, because
103=1000.10^3=1000.


More generally, if
x=byx=b^y
, then
yy
is the logarithm base
bb
of
xx
, written:
y=logb(x)y=\log_b(x)
, so
log10(1000)=3\log_{10}(1000)=3
.

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

logb(bz)=z\displaystyle \log_b(b^z)=z


And:

blogb(z)=z\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:

logb(xy)=logb(x)+logb(y)\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
logb(x)=v\log_b(x)=v
 and
logb(y)=w.\log_b(y)=w.


Writing these equations as exponentials:

bv=x\displaystyle b^v=x


And:

bw=y.\displaystyle b^w=y.


Then note that:

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


Taking the logarithm base
bb
 of both sides of this last equation yields:

logb(xy)=logb(bv+w)=v+w=logb(x)+logb(y)\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:

log10(10x100x3+1)=log10(10x)+log10(100x3+1)\displaystyle \log_{10}(10^x\cdot 100^{x^3+1})=\log_{10} (10^x)+\log_{10}(100^{x^3+1})


Then because
100100
 is
10210^2
, we have:

x+log10(102(x3+1))=x+2(x3+1)=2x3+x+2\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
pthp\text{th}
power of a quantity is
pp
times the logarithm of the quantity. In symbols, 
logb(xp)=plogb(x).\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,
    pp
    , signifies that a number is being multiplied by itself
    pp
    number of times. Because the logarithm of a product is the sum of the logarithms of the factors, the logarithm of a number,
    xx
    , to an exponent,
    pp
    , is the same as the logarithm of
    xx
     added together
    pp
    times, so it is equal to 
    plogb(x).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
    x3x^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:

logb(xy)=logb(x)+logb(y)\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
logbxp\log_b x^p
.

Recall that
xpx^p
 can be thought of as
xxxxx \cdot x \cdot x \cdots x
 where there are
pp
 factors of
xx
. Then we have:

logb(xp)=logb(xxx)=logbx+logbx++logbx=plogbx\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
pp
 factors of
xx
 are converted to
pp
 summands by the product rule formula.

Example 1: Simplify the expression 
log3(3x9x100)\log_3(3^x\cdot 9x^{100})

First expand the log:

log3(3x9x100)=log3(3x)+log39+log3(x100)\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:

log3(3x)+log39+log3(x100)=x+2+100log3x\displaystyle \log_3 (3^x) + \log_3 9 + \log_3 (x^{100})= x+2+100\log_3 x


Example 2: Solve
2(x+1)=1032^{(x+1)}=10^3
 for
xx
 using logarithms

Start by taking the logarithm with base
22
 of both sides:

log2(2(x+1))=log2(103)x+1=3log2(10)x=3log2(10)1\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=3log2(10)1.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,
logb(xy)=logbxlogby.\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
    x3x^3
    .


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

logb(xy)=logbx+logby\displaystyle \log_b(xy) = \log_bx + \log_by


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

logb(xy)=logbxlogby.\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=logbxu=\log_b x
 and
v=logbyv=\log_b y
.

Then
bu=xb^u=x
 and
bv=y.b^v=y.


Then:

logb(xy)=logb(bubv)=logb(buv)=uv=logbxlogby\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
yy
 is the same is multiplying by
y1.y^{-1}.
 So we can write:

logb(xy)=logb(xy1)=logbx+logb(y1)=logbxlogby\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
log2(x4y9z100)\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:

log2(x4)+log2(y9)log2(z100)=4log2x+9log2y100log2z.\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
    1010
    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
1010
, but do not have keys for other bases. So, if you needed to get an approximation to a number like
log4(9)\log_4(9)
 it can be difficult to do so. One could easily guess that it is between
11
 and
22
 since
99
 is between
414^1
 and
424^2
, 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:

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


Thus, for example, we could calculate that
log4(9)=log10(9)log10(4)\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
loga(x)\log_a(x)
 a name like
zz
:

z=loga(x)\displaystyle z=\log_a(x)


Write this as 
az=xa^z=x


Now take the logarithm with base
bb
 of both sides, yielding:

logbaz=logbx\displaystyle \log_b a^z = \log_bx


Using the power rule gives:

zlogba=logbx\displaystyle z \cdot \log_ba = \log_b x


Dividing both sides by
logba\log_ba
 gives:

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


Thus we have 
logax=logbxlogba.\log_a x ={\log_b x \over \log_b a}.


Example

An expression of the form
log5(10x2+1)\log_5(10^{x^2+1})
 might be easier to graph on a graphing calculator or other device if it were written in base
1010
 instead of base 5. The change-of-base formula can be applied to it:

log5(10x2+1)=log10(10x2+1)log105\displaystyle \log_5(10^{x^2+1}) = {\log_{10}(10^{x^2+1}) \over \log_{10}5}


Which can be written as 
x2+1log105.{x^2+1 \over \log_{10} 5}.


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