A logarithmic equation has at least one variable inside a logarithm.
For example, ‘$\,\ln(3x1) = 5\,$’ 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’:
In particular, any equation that can be put in the form ‘$\,\log_b(\text{stuff}) = \text{constant}\,$’ can be solved using IUSC, where:
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):
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:
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 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 $\log(x + 2) + \log(x1) = 1$.
The first step in IUSC (Isolate) requires that we get a single logarithm.
To this end, the expression ‘$\,\log(x+2) + \log(x1)\,$’ is rewritten as ‘$\,\log(x+2)(x1)\,$’.
(Recall that the log of a product is the sum of the logs.)
This simple rewriting is where the problem can occur, as follows.
Recall that logarithms only act on strictly positive numbers.
Therefore, in the expression ‘$\,\log(x + 2) + \log(x1)\,$’ we must have both
$\,x + 2 > 0\,$ and $\,x  1 > 0\,$.
However, in the expression ‘$\,\log(x+2)(x1)\,$’ both $\,x+2\,$ and
$\,x1\,$ might be negative,
yielding an acceptable positive product $\,(x+2)(x1)\,$.
For example, let $\,x = 3\,$.
For this value of $\,x\,$:
Any value of $\,x\,$ for which the equation ‘$\log(x + 2) + \log(x1) = 1$’ is undefined,
but for which the equation ‘$\log(x + 2)(x1) = 1$’ is true,
would be an extraneous solution.
Study the second example below to see what happens for this particular equation.
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, cutandpaste the following at WolframAlpha: 
Solve: $\log(x+2) + \log(x1) = 1$
$\log(x+2) + \log(x1) = 1$  (original equation) 
$\log(x+2)(x1) = 1\qquad \bigstar$ 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. 
$10^1 = (x+2)(x1)$ 
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. 
$(x+4)(x3) = 0$ 
Solve (continued): Factor the LHS. 
$x = 4$ or $x = 3$ 
Solve (continued): Use the Zero Factor Law. There are two potential solutions. 
$\log(4+2) + \log(41) \overset{\text{?}}{=} 1$ $\log(2) + \log(5)\overset{\text{?}}{=} 1$ 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. 
$\log(3+2) + \log(31) \overset{\text{?}}{=} 1$ $\log(5) + \log(2)\overset{\text{?}}{=} 1$ $1=1$ Okay! 
Check second potential solution: There is a unique solution for this equation. 
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
