12.7 - Logarithmic equations
Before reading this section you may want to review the following topics:
A logarithmic equation is one where the variable to be solved for (call it x)
is in the argument of a logarithm function.
For example log(x) = 5 is a logarithmic equation while log(5) = x is not.
To solve a logarithmic equation follow these steps:
- If the equation contains several logarithms then you must first use the
properties of logarithms
to combine them into a single logarithm.
- Isolate the logarithm. This means put the equation into the form
log b( f ) = a so that the log
function is alone on one side of the equation. (The expression
f contains the unknown x.)
- Antilog both sides, thus putting the
equation into the exponential form b a = f.
- The unknown x is no longer inside a logarithm.
Now you can finish solving for x
by using the basic procedures for solving equations.
- Check the solution.
Example: Solve the logarithmic equation
2 log 3 (x − 1) = 4 for x.
Solution: There is only a single logarithm so the first step is to isolate the
logarithm. To do this divide both sides by 2:
log 3 (x − 1) = 2
The next step is to antilog both sides:
x − 1 = 3 2
The equation is no longer logarithmic and we can finish solving for x
by simply adding 1 to both sides:
x = 10
Example: Solve the logarithmic equation
log (3 x + 1) − 2 log (x) = 1 for x:
Solution: The first step is to combine the logarithms using the
properties of logarithms.
First use property 3, then property 2:
log(3 x + 1) − log (x2 ) = 1
The next step is to antilog both sides.
Note that the base of the logarithm is understood to be 10.
The equation is no longer logarithmic - it is fractional, so we can proceed to solve for x
using techniques for fractional equations. Clear the denominator
by multiplying through by x2 and then move all terms to the
left side:
10 x 2 − 3 x − 1 = 0
The result is a quadratic equation in standard form.
The left-hand-side can be factored:
(5 x + 1) (2 x − 1) = 0.
We can replace this equation by two new equations, each of which results from setting
one of the factors equal to zero. Solving them yields the solutions:
Now we must check the solutions. Substituting x = 1/2 into the original
equation and simplifying yields the equation 1 = 1, so it checks out. But substituting
x = −1/5 into the original equation means that we must evaluate the logarithm of
a negative number and this cannot be done over the real numbers. Thus this solution
is extraneous; which leaves us with the only solution, x = 1/2.
Note that not all logarithmic equations can be solved using algebra.
For example consider the seemingly simple equation x = log (x).
We cannot get the x out of the logarithm without putting the other x
into an exponential. This equation can only be solved approximately using a computer.