## SOLVING LOGARITHMIC EQUATIONS

Note:

If you would like an in-depth review of logarithms, the rules of logarithms, logarithmic functions and logarithmic equations, click on logarithmic function.

Solve for x in the following equation.

Example 3:

The above equation is valid only if all the terms in the equation are valid. The term is valid if the term is valid if and the term is valid if The domain is the set of real numbers that are greater than , greater than 10, and greater than -2. The domain is therefore the set of real numbers greater than 10.

Algebraic Method:

Although there are two solutions, only one of the solutions is in the domain of the problem. The exact solution to the problem is , and the approximate solution is

Graphing Method:

Change the original equation into an equation where all the logarithmic terms have base 10.

Rewrite the equation as

Let's call the left side of the equation f(x) and the right side of the equation g(x).

Then and Graph f(x) and g(x). We are looking for the point(s) of intersection, The solution, if any, will be the value of x in the point(s) of intersection.

The graph of the right side of the equation is the set of points where the value of y equals zero. This is easy; it is the x-axis. We then look to see where the graph of f(x) crosses the x-axis.

Note that the graph only appears to the right of This is consistent with our finding that the domain of the original equation is the set of real numbers greater than .

The solution(s) to the original equation is the set of real numbers where f(x) crosses the x-axis (the x-intercepts are the solutions to the problem.) . You will note from the graph that f(x) crosses the x-axis at x=1.6735161761. This means that the equation has one real solution at .

Algebraic Method:

Change the bases of the logarithmic terms to 10.

Multiply both sides of the equation by the factor

An algebraic solution, other than interpolation, is too difficult for a beginning student.

Interpolation Method:

Since the domain is the set of real numbers greater than choose numbers larger than

• Let's start with x=1.6and evaluate f(1.6). Since we are looking for the value such that our target is 0.

• Let's now try x=10 and evaluate f(10). Since we are looking for the value such that our target is 0.

Since f(1.6)<0 and f(10)>0, we know that the graph of f(x) must cross the x-axis between x=1.6 and x=10. This means that we know at least one solution has a value symbol> than 1.6 and less than 10. Since is closer to zero than , we know that the answer most likely lies to the left of the midpoint of 1.6 and 10, or to the left of 7.4.

• Let's try x=4.5 (4.5 is located to the left of 7.4) and evaluate f(4.5). We are looking for a value of x such that our target is 0.

Since f(1.6)<0 and f(4.5)>0, we know that the graph of f(x) must cross the x-axis between x=1.6 and x=4.5. This means that we know at least one solution has a value greater than 1.6 and less than 4.5. Since is closer to zero than , we know that the answer most likely lies to the left of the midpoint of 1.6 and 4.5 or to the left of 3.05..

• Let's try x=2 and evaluate f(2). Since we are looking for the value such that our target is 0.

Since f(1.6)<0 and f(2)>0, we know that the graph of f(x) must cross the x-axis between x=1.6 and x=2. This means that we know at least one solution has a value greater than 1.6 and less than 2. Since is closer to zero than , we know that the answer most likely lies to the left of the midpoint of 1.6 and 2 or to the left of 1.8.

• Let's try x=1.7 and evaluate f(1.7). Since we are looking for the value such that our target is 0.

Since f(1.6)<0 and f(1.7)>0, we know that the graph of f(x) must cross the x-axis between x=1.6 and x=1.7. This means that we know at least one solution has a value greater than 1.6 and less than 1.7. Since is closer to zero than the answer is closer to 1.7 than to 1.6.

• Let's try x=1.68 and evaluate f(1.68). Since we are looking for the value such that our target is 0.

Since f(1.6)<0 and f(1.68)>0, we know that the graph of f(x) must cross the x-axis between x=1.6 and x=1.68. This means that we know at least one solution has a value greater than 1.6 and less than 1.68. Since is closer to zero than the answer is closer to 1.68 than to 1.6.

• Let's try x=1.67 and evaluate f(1.67). Since we are looking for the value such that our target is 0.

Since f(1.67)<0 and f(1.68)>0, we know that the graph of f(x) must cross the x-axis between x=1.67 and x=1.68. This means that we know at least one solution has a value greater than 1.67 and less than 1.68.

• You can now try x=1.675 and place the solution between x=1.67 and x=1.675 or between x=1.675 and x=1.68. You con continue in this fashion until your approximation of the answer is carried to as many decimals as your problems requires. Let's try x=1.68= and evaluate f(1.68). Since we are looking for the value such that our target is 0.

Since f(1.6)<0 and f(1.68)>0, we know that the graph of f(x) must cross the x-axis between x=1.6 and x=1.68. This means that we know at least one solution has a value greater than 1.6 and less than 1.68. Since is closer to zero than the answer is closer to 1.68 than to 1.6.

If you would like to test yourself by working some problems similar to this example, click on problem.

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Author: Nancy Marcus