Solving Logarithmic Equations

A logarithmic equation has at least one variable inside a logarithm.

For example, ‘$\,\ln(3x-1) = 5\,$’ is a logarithmic equation, since there is a variable $\,x\,$ inside the logarithm.
However, ‘$x\log 3 = 5$’ is not a logarithmic equation, since there is no variable inside a logarithm.

As with solving exponential equations, many logarithmic equations can be solved by a technique that can be abbreviated as ‘IUSC’:

Isolate $\ldots$ Undo with an exponential function $\ldots$ Solve resulting equation $\ldots$ Check

In particular, any equation that can be put in the form ‘$\,\log_b(\text{stuff}) = \text{constant}\,$’ can be solved using IUSC, where:

$b > 0\,$ and $\,b \ne 1\,$ (the allowable bases for logarithms)
‘stuff’ is linear ($\,ax+b\,$) or quadratic ($\,ax^2 + bx + c\,$)

The IUSC method for solving logarithmic equations may introduce extraneous solutions. Extraneous solutions are discussed below.

Here are examples of logarithmic equations solvable by IUSC (and each is solved below):

$4 + 3\log(2x) = 16$
$\log(x + 2) + \log(x-1) = 1$

Extraneous Solutions

Some things that can ‘be done’ to an equation may NOT result in an equivalent equation!
When solving equations, you should try to avoid such things, and stick with actions that are certain to yield equivalent equations.
Sometimes, however, this is not possible—and you must remain aware that you may have lost or added a solution.

For example, squaring both sides of an equation can take a false statement to a true statement:

$$ \begin{alignat}{2} 2\ \ &=\ -2 \qquad\qquad&&\text{(false)}\cr 2^2\ &=\ (-2)^2 &&\text{(square both sides)}\cr 4\ \ &=\ 4 &&\text{(true)} \end{alignat} $$

When variables are involved, this can cause a solution to be added:

Consider the equation ‘$\,x = 2\,$’, which has solution set $\,\{2\}\,.$
Squaring both sides of ‘$\,x = 2\,$’ yields the new equation ‘$\,x^2 = 4\,$’.
The solution set of ‘$\,x^2 = 4\,$’ is $\,\{2,-2\}\,.$
Thus, squaring both sides caused a solution to be added.

Definition extraneous solutions

An extraneous solution is a number for which an original equation is either undefined or false,
but for which a later equation is true.

Extraneous solutions are caused by actions that can take a false (or undefined) equation to a true equation.

Extraneous solutions are discarded, since they are not solutions of the original equation.

It's a good idea to put a ‘$\,\bigstar\,$’ next to any step in a solution process that has the potential of producing an extraneous solution.
The ‘$\,\bigstar\,$’ is a reminder to yourself that you may have added a solution.
In such cases, you must check the potential solution(s), and discard any that don't make the original equation true.
The example below illustrates the process.

Extraneous Solutions When Solving Logarithmic Equations

Extraneous solutions can occur when solving logarithmic equations, so be careful!
The problem typically arises when properties of logarithms are used to combine two or more logarithmic terms into a single term, as discussed next.

Consider solving the equation $\log(x + 2) + \log(x-1) = 1$.

The first step in IUSC (Isolate) requires that we get a single logarithm.
To this end, the expression ‘$\,\log(x+2) + \log(x-1)\,$’ is rewritten as ‘$\,\log[(x+2)(x-1)]\,$’.
(Recall that the log of a product is the sum of the logs.)
This simple re-writing is where the problem can occur, as follows.

Recall that logarithms only act on strictly positive numbers.
Therefore, in the expression ‘$\,\log(x + 2) + \log(x-1)\,$’ we must have both $\,x + 2 > 0\,$ and $\,x - 1 > 0\,.$
However, in the expression ‘$\,\log[(x+2)(x-1)]\,$’ both $\,x+2\,$ and $\,x-1\,$ might be negative,
yielding an acceptable positive product $\,(x+2)(x-1)\,.$

For example, let $\,x = -3\,.$ For this value of $\,x\,$:

$\log (x + 2) = \log (-3 + 2) = \log(-1)\,$ is not defined, since logs cannot act on negative numbers
$\log (x -1) = \log (-3 -1) = \log(-4)\,$ is not defined, since logs cannot act on negative numbers
$\log[(x + 2)(x-1)] = \log[(-3+2)(-3-1)] = \log[(-1)(-4)] = \log 8\,$ is defined

Any value of $\,x\,$ for which the equation ‘$\log(x + 2) + \log(x-1) = 1$’ is undefined,
but for which the equation ‘$\log[(x + 2)(x-1)] = 1$’ is true, would be an extraneous solution.
Study the second example below to see what happens for this particular equation.

EXAMPLE

Solve: $4 + 3\log(2x) = 16$

$4 + 3\log(2x) = 16$	(original equation)
$3\log(2x) = 12$	Isolate: Subtract $\,4\,$ from both sides.
$\log(2x) = 4$	Isolate (continued): Divide both sides by $\,3\,.$
$2x = 10^4$	Undo: Write the equivalent exponential equation. (Equivalently, you can think of raising $\,10\,$ to both sides.) Recall that ‘$\log$’ (without any indicated base) means log base ten (the common logarithm).
$\displaystyle x = \frac{10^4}{2} = 5000$	Solve: Divide both sides by $\,2\,.$
$4 + 3\log(2\cdot 5000) \overset{\text{?}}{=} 16$ $16 = 16$ Okay!	Check: At WolframAlpha, ‘$\,\log\,$’ (with no indicated base) means the natural logarithm. To get the common logarithm, you can use ‘`log_{10}`’. For example, cut-and-paste the following at WolframAlpha: `4 + 3log_{10}(2*5000)`

EXAMPLE

Solve: $\log(x+2) + \log(x-1) = 1$

$\log(x+2) + \log(x-1) = 1$	(original equation)
$\log[(x+2)(x-1)] = 1\qquad \bigstar$ The $\,\bigstar\,$ indicates that there is the potential for an extraneous solution at this step. You must check all potential solution(s) at the final step, to see if they make the original equation true.	Isolate: Use a property of logs—the log of a product is the sum of the logs—to get a single logarithmic term.
$10^1 = (x+2)(x-1)$	Undo: Write the equivalent exponential equation. This is now a quadratic equation, which will next be put in standard form.
$x^2 + x - 2 = 10$	Solve: Multiply out, and switch sides.
$x^2 + x - 12 = 0$	Solve (continued): Subtract $\,10\,$ from both sides. The equation is now in standard form.
$(x+4)(x-3) = 0$	Solve (continued): Factor the LHS.
$x = -4$ or $x = 3$	Solve (continued): Use the Zero Factor Law. There are two potential solutions.
$\log(-4+2) + \log(-4-1) \overset{\text{?}}{=} 1$ $\log(-2) + \log(-5)\overset{\text{?}}{=} 1$ No! $-4\,$ is an extraneous solution	Check first potential solution: Since logs cannot act on negative numbers, $\,-4\,$ is an extraneous solution, and is discarded.
$\log(3+2) + \log(3-1) \overset{\text{?}}{=} 1$ $\log(5) + \log(2)\overset{\text{?}}{=} 1$ $1=1$ Okay!	Check second potential solution: There is a unique solution for this equation.

Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Solving Exponential Growth and Decay Problems

SOLVING LOGARITHMIC EQUATIONS

Extraneous Solutions

Extraneous Solutions When Solving Logarithmic Equations

EXAMPLE

EXAMPLE