### All Precalculus Resources

## Example Questions

### Example Question #1 : Properties Of Logarithms

Solve for :

**Possible Answers:**

The correct solution set is not included among the other choices.

**Correct answer:**

The correct solution set is not included among the other choices.

FOIL:

These are our *possible* solutions. However, we need to test them.

:

The equation becomes . This is true, so is a solution.

:

However, negative numbers do not have logarithms, so this equation is meaningless. is not a solution, and is the one and only solution. Since this is not one of our choices, the correct response is "The correct solution set is not included among the other choices."

### Example Question #1 : Properties Of Logarithms

Solve for x:

**Possible Answers:**

**Correct answer:**

The key to simplifying this problem is by using the Natural Logarithm Quotient Rule .

Using algebraic manipulation to bring each natural logarithm to one side, we obtain:

### Example Question #1 : Properties Of Logarithms

Solve for x:

**Possible Answers:**

and

and

**Correct answer:**

Using the logarithmic product rule , we simplify as follows:

Factoring this quadratic equation, we will obtain two roots.

The roots are and . However, the domain of the logarithmic function is . That is to say, it is not defined for numbers less than or equal to 0. So our final answer is

### Example Question #1 : Properties Of Logarithms

Simplify the expression as a single natural logarithm with a coefficient of one: .

**Possible Answers:**

**Correct answer:**

Here we need to make use the power rule

Recall that , so we have

### Example Question #1 : Properties Of Logarithms

Evalute the equation .

**Possible Answers:**

**Correct answer:**

We can rewrite as , and then multiply each side by . Doing so, we get

This is just a quadratic equation with replacing . Let us factor it just like a quadratic equation.

in this case is a root with multiplicity of two, so there are two answers to this equality, both of them being .

### Example Question #1 : Properties Of Logarithms

Evalute the equation . Use a calculator to approximate the answer to three decimal places.

**Possible Answers:**

**Correct answer:**

In order to evaluate this equation, we have to do some algebraic manipulation first to get the exponential function isolated.

### Example Question #1 : Properties Of Logarithms

solve for x: .

**Possible Answers:**

**Correct answer:**

Here we employ the use of the logarithm base change formula .

We could convert either or to the other's base. Let's convert to a logarithm with base 4.

Plugging this back in to the original equation,

### Example Question #1 : Properties Of Logarithms

Calculators are not requried (and are strongly discouraged) for this problem.

Expand and simplify the following logarithm:

**Possible Answers:**

**Correct answer:**

First expand the logarithm using the product property:

We can evaluate the constant log on the left either by memorization, sight inspection, or deliberately re-writing 16 as a power of 4, which we will show here:

, so our expression becomes:

Now use the power property of logarithms:

Rewrite the equation accordingly.

Note that the 3rd terms becomes negative because the exponent is negative. We will use one last log property to finish simplifying:

Accordingly,

.

Now substitute and simplify:

### Example Question #1 : Properties Of Logarithms

Simplify:

**Possible Answers:**

**Correct answer:**

First use the reversal of the logarithm power property to bring coefficients of the logs back inside the arguments:

Now apply this rule to every log in the formula and simplify:

Next, use a reversal of the change-of-base theorem to collapse the quotient:

Substituting, we get:

Now combine the two using the reversal of the logarithm product property:

### Example Question #2 : Properties Of Logarithms

Find the inverse function of the following exponential function:

**Possible Answers:**

**Correct answer:**

Since we are looking for an inverse function, we start by swapping the x and y variables in our original equation.

Now we have to solve for y. To do this we have to work towards isolating y. First we remove the constant multiplier:

Next we eliminate the base on the right side by taking the natural log of both sides.

Subtract 1 and divide by 4:

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