3.4: Solving Logarithmic Equations
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In the previous section, we took exponential equations and used the properties of logarithms to restate them as logarithmic equations. In this section, we will take logarithmic equations and use properties of logarithms to restate them as exponential equations. In the previous section, we used the property of logarithms that said logbMp=plogbM. In this section, we will make use of two additional properties of logarithms:
logb(M∗N)=logbM+logbN
and
logbMN=logbM−logbN
Just as our previous property of logarithms was simply a restatement of the rules of expoenents, these two properties of logarithms depend on the rules of exponents as well. since we're interested in logbM and logbN, let's restate these in terms of exponents:
If logbM=x then bx=M and if logbN=y then by=N
The properties of logarithms we're interested in justifying have to do with M∗N and MN, so let's look at those expressions in terms of exponents:
M∗N=bx∗by=bx+y and MN=bxby=bx−y
If we're interested in logb(M∗N), then we're asking the question "What power do we raise b to in order to get M∗N?" We can see above that raising b to the x+y power gives us M∗N. since x=logbM and y=logbN then x+y= logbM+logbN, so:
logb(M∗N)=x+y=logbM+logbN
Likewise, if we're interested in logbMN, we're asking the question "What power do we raise b to in order to get MN?" since raising b to the x−y power gives us MN and x−y=logbM−logbN, then:
logbMN=x−y=logbM−logbN
Let's look at an example to see how we'll use this to solve equations:
Example
Solve for x
log2x+log2(x−4)=2
The first thing we can do here is to combine the two logarithmic statements into one. since logb(M∗N)=logbM+logbN, then log2x+log2(x−4)=log2[x(x−4)]
log2x+log2(x−4)=2log2[x(x−4)]=2
Then we'll restate the resulting logarithmic relationship as an exponential relationship:
22=x(x−4)4=x2−4x0=x2−4x−44.828,−0.828≈x
Most textbooks reject answers that result in taking the logarithm of a negative number, such as would be the case for x≈−0.828. However, the logarithms of negative numbers result in complex valued answers, rather than an undefined quantity. For that reason, in this text, we will include all answers.
If a problem involves a difference of logarithms, we can use the other property of logarithms introduced in this section.
Example
Solve for x
log(5x−1)−log(x−2)=2
Again, our first step is to restate the difference of logarithms using the property logbMN=logbM−logbN:
log(5x−1)−log(x−2)=2log[5x−1x−2]=2
We're working with a logarithm in base 10 in this problem, so in our next step we'll say:
log[5x−1x−2]=25x−1x−2=102
Then multiply on both sides by x−2
102=5x−1x−2(x−2)∗100=5x−1(x−2)∗(x−2)
And, solve for x
100x−200=5x−195x=199x=19995
In some equations, all of the terms are stated using logartihms. These equations often come out in a form that says logbx=logby. If this is the case, we can then conclude that x=y
It seems reasonable that if the exponent we raise b to in order to get x is the same exponent that we raise b to in order to get y, then x and y are the same thing.
Assume:
logbx=logby
then
ba=x and ba=y
if both x and y are equal to ba, then x=y
Example
Solve for x
log5(4−x)=log5(x+8)+log5(2x+13)
First, let's use the properties of logarithms to restate the equation so that there is only one logarithm on each side.
log5(4−x)=log5(x+8)+log5(2x+13)log5(4−x)=log5[(x+8)(2x+13)]
Then, we'll use the property of logarithms we just discussed:
If logbx=logby
then
x=ylog5(4−x)=log5[(x+8)(2x+13)]4−x=(x+8)(2x+13)0=2x2+29x+1040=2(x+5)(x+10)−5,−10=x
Exercises 3.4
Solve for the indicated variable in each equation.
1) log35+log3x=2
2) log4x+log45=1
3) log2x=2+log23
4) log5x=2+log53
5) log3x+log3(x−8)=2
6) log6x+log6(x−5)=1
7) log(3x+2)=log(x−4)+1
8) log(x−1)−logx=−0.5
9) log2a+log2(a+2)=3
10) log3x+log3(x−2)=1
11) log2y−log2(y−2)=3
12) log2x−log2(x+3)=2
13) log3x+log3(x+4)=2
14) log4u+log4(u+1)=1
15) log5+logx=log6
16) lnx+ln4=ln2
17) log7x−log712=log72
18) log2−logx=log8
19) log3x−log3(x−2)=log34
20) log62−log6(x−2)=log69
21) log4x−log4(x−4)=log4(x−6)
22) log9(2x+7)−log9(x−1)=log9(x−7)
23) 2log2x=log2(2x−1)
24) 2log4y=log4(y+2)
25) 2log(x−3)−3log2=1
26) 2log57−log5(x+1)=log5(2x−5)