Solving Logarithmic Equations
Now that we have the following rules, we can use them to solve equations.
logb x = n iff bn = x
log ab = log a + log b.
log a/b = log a – log b
log an = n log a
logbn = logan/logab
So you need to remember, there are two types of logarithmic problems; log equals
number and log = log.
1.
If log = #, then we use logba = n if and only if a = bn and solve
2.
If log = log, then we use logbx1 = logbx2 iff x1 = x2 and solve
But to use those, you first have to simplify the logarithmic expressions using the
product, quotient, power or change of base rules.
Our initial job is to rewrite the exponential or logarithmic equations into one of
those two forms using the rules we derived.
Example
Solve for x,
log x + log (x–3) = 1
Using the product rule
log x(x–3) = 1
Using the definition and knowing when a base is not written it is
understood to be 10, we have
x(x–3) = 101
Using the D-Prop
x2 – 3x = 10
Solving for x
x2 – 3x – 10 = 0
(x+2)(x–5) = 0
x + 2 = 0 or x – 5 =0
x=–2
or x = 5
***Important, you must check your answers! You can only take a log
of a positive number. If x = –2, then we would be taking a log of a
negative number – that can not be a solution. The answer is x = 5
Example
Solve for x,
log2(x + 8) + log2(x – 4) = 3
Using the product rule
log2(x + 8)(x – 4) = 3
Using the definition
(x + 8)(x – 4) = 23
Multiplying
x2 + 4x – 32 = 8
Solving for x
x2 + 4x – 40 = 0
Use Quadratic Formula a = 1, b = 4, c = –40
x=
−(4) ± 4 2 − 4(1)(–40
2(1)
−4 ± 16 + 160
2
−4 ± 176
x=
2
−4 ± 13.2
x=
2
x = 4.6
x = −8.6
x=
x can not be –8.6 because we can not take the log of a negative number,
4.6 is the solution.
x=
Now, even though this problem took more steps to solve, that should not equate to
this problem being more difficult. Rather than solving it by factoring and using the
Zero Product Property, we used the Quadratic Formula.
Now, let’s look at a problem where we have logs on both sides of the equation.
Remember the rule, once the equation is in simplified form, we drop the logs.
Example
Solve for x,
Using the product and exp rules
log(x–2) + log(2x–3) = 2logx
log (x–2)(2x–3) = logx2
(x – 2)(2x – 3) = x2
2x2 – 7x + 6 = x2
x2 – 7x + 6 = 0
(x – 6)(x – 1) = 0
x=6
or
x=1
When x = 1, results in taking a log of a negative number – can’t happen!
x=6
Example
Solve for x,
log 4x – log4(x – 1) = ½
Using the quotient rule
log4
x
=½
x −1
Using the definition
x
x −1
=4½
x
=2
x −1
2(x – 1) = x
2x – 2 = x
x=2
Checking the answer, x = 2 works.