Solving and Graphing Logarithms - Algebra 2

Set and solve:

The -intercept is .

Set and solve:

The -intercept is .

The graph of the logarithmic function

has as its only asymptote the vertical line

Here, since , the only asymptote is the line

.

There is no real number for which

Therefore in the equation , cannot be

However, can be infinitely large or negative.

Finally, when or twice as large.

is the inverse of and therefore the graph is simply the mirror image flipped over the line

Let . In terms of ,

.

The graph of has as its vertical asymptote the line of the equation . The graph of is the result of three transformations on the graph of - a left shift of 4 units , a vertical stretch ( ), and an upward shift of 2 units ( ). Of the three transformations, only the left shift affects the position of the vertical asymptote - the asymptote of also shifts left 4 units, to .

Let

In terms of ,

This is the graph of shifted left 4 units, stretched vertically by a factor of 3, then shifted up 2 units.

The graph of does not have a horizontal asymptote; therefore, a transformation of this graph, such as that of , does not have a horizontal asymptote either.

To solve for , first convert both sides to the same base:

Now, with the same base, the exponents can be set equal to each other:

Solving for gives:

Rewrite in exponential form:

Solve for x:

Logs are exponential functions using base 10 and a property is that you can combine added logs by multiplying.

You cannot take the log of a negative number. x=-25 is extraneous.

Rewrite as a product that includes the number :

Then we can split up the logarithm using the Product Property of Logarithms:

Thus,

.

To solve this logarithm, we need to know how to read a logarithm. A logarithm is the inverse of an exponential function. If a exponential equation is

then its inverse function, or logarithm, is

Therefore, for this problem, in order to solve for , we simply need to solve

which is .

For this problem it is helpful to remember that,

is equivalent to because

Therefore we can set what is inside of the parentheses equal to each other and solve for as follows:

This question is testing the definition of logs. is the same as .

In this case, can be rewritten as .

Taking square roots of both sides, you get . Since only the positive answer is one of the answer choices, is the correct answer.

The first step is to rewrite this equation in log form.

When rewriting an exponential function as a log we must remember that the form of an exponential is:

When this is rewritten in log form it is:

.

Therefore we have which when rewritten gives us,

.

Logarithms are another way of writing exponents. In the general case, really just means . We take the base of the logarithm (in our case, 2), raise it to whatever is on the other side of the equal sign (in our case, 4) and set that equal to what is inside the parentheses of the logarithm (in our case, x+6). Translating, we convert our original logarithm equation into . The left side of the equation yields 16, thus .

To solve this equation, remember log rules

.

This rule can be applied here so that

and

Because both sides have the same logarithmic base, both terms can be set equal to each other:

Now, evaluate the equation.

First, add x to both sides:

Add 15 to both sides:

Finally, divide by 6: .

In logarithmic expressions, is the same thing as .

Therefore, the equation can be rewritten as .

Both 8 and 128 are powers of 2, so the equation can then be rewritten as .

Since both sides have the same base, set .

Solve by dividing both sides of the equation by 3: .