### 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 bf ) = 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 (x) = 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 x 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.

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