### All High School Math Resources

## Example Questions

### Example Question #11 : Solving Logarithmic Equations

Solve the equation.

**Possible Answers:**

No solution

**Correct answer:**

No solution

Change 125 to so that both sides have the same base. Apply log and then set the exponential expressions equal to each other so that . Upon trying to isolate , it becomes clear that there is no solution.

### Example Question #12 : Solving Logarithmic Equations

Solve for :

**Possible Answers:**

The equation has no solution.

**Correct answer:**

Since , , and we can rewrite and solve this statement as follows:

Substitution confirms this to be the only solution.

### Example Question #13 : Solving Logarithmic Equations

Solve for :

**Possible Answers:**

The equation has no solution.

The solution set of the equation is the set of all real numbers.

**Correct answer:**

We can rewrite this as follows:

Split this into a compound statement, and solve:

Test both of these possible solutions by substituting:

Since only positive numbers can have logarithms, this equation does not make sense. is not a solution.

is a solution - the only solution.

### Example Question #14 : Solving Logarithmic Equations

What is the sum of the value(s) of x that satisfy the equation ?

**Possible Answers:**

20

100

No solution.

35/2

35

**Correct answer:**

35/2

We need to rewrite the logarithmic equation into a form that is more recognizable.

An equation in the form can be rewritten in the exponential form , where a, b, and c are constants (and b > 0). Thus, we will rewrite our original equation as follows:

.

We can now approach this as we would a typical quadratic equation.

Subtract 100 from both sides. We need to set all of the terms equal to zero.

.

We can factor this by first multiplying the outer coefficients. The product of 2 and -100 is equal to -200. We must think of two numbers that multiply to give -200 but add to give -35 (the middle coefficient). Two numbers that satisfy both of these requirements are -40 and 5. We will now rewrite the quadratic polynomial.

We can use grouping to factor the polynomial. Factor the first two terms and the last two terms separately.

Notice that we can now factor out x - 20 from the 2x term and the 5 term.

To solve this, we set each factor equal to zero.

The question ultimately asks for the sum of the values of x that satisfy the equation, so we must add -5/2 and 20, which yields 35/2.

The answer is 35/2.

### Example Question #15 : Solving Logarithmic Equations

Solve the following logarithmic equation:

**Possible Answers:**

**Correct answer:**

Since the bases of the logs are the same and the logarithms are added, the arguments can be multiplied together.

The logarithm can then be converted to exponential form.

Although there are two solutions to the equation, logarithims cannot be negative. Therefore, the only real solution is .

### Example Question #16 : Solving Logarithmic Equations

Solve the following logarithmic equation:

**Possible Answers:**

**Correct answer:**

Start by converting the logarithms into exponential form:

Convert again into exponential form:

Simplify and solve:

### Example Question #17 : Solving Logarithmic Equations

Solve the following logarithmic equation:

**Possible Answers:**

**Correct answer:**

Since the bases of the logs are the same, the terms inside the logs can be set equal to each other.

### Example Question #18 : Solving Logarithmic Equations

Solve the equation.

**Possible Answers:**

**Correct answer:**

Change the right side to so that both sides are the same. Apply log to both sides so that you can set the exponential expressions equal to each other ().

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