Unit 2 Notes: Understanding Logarithms (AP Precalculus)

Logarithmic Expressions and Functions

What a logarithm means (and why you should care)

A logarithm answers a very specific question:

“What exponent do I need?”

You already know how exponential functions work: you start with a base and raise it to some exponent. Logarithms reverse that process. For example, if you know that

2^5 = 32

then the logarithm tells you the exponent 5 when the base is 2 and the result is 32.

Formally, the logarithm base b of a positive number x is defined by the equivalence:

\log_b(x) = y \text{ if and only if } b^y = x

This definition matters because many real situations involve repeated multiplication (growth/decay, compounding, sound intensity, acidity). When you need the time or exponent and not the final amount, logarithms are the natural tool.

Conditions you must remember (domain and base restrictions)

The definition above only makes sense under certain conditions.

• The base must satisfy:

b > 0

b \ne 1

• The input (the “argument”) must satisfy:

x > 0

That last line is a big deal for graphs and for solving equations: you can’t take the logarithm of zero or a negative number in the real number system.

Logarithmic expressions and notation

A logarithmic expression is any expression built using logs, such as:

\log_3(7x - 1)

or

2\log(x) - \log(5)

You will see multiple notations:

The natural logarithm uses the constant e as its base. The constant e is the standard base for continuous growth models and appears throughout calculus, but in precalculus you mainly use it as “another common base” with its own convenient button on calculators.

The logarithmic function as an inverse

The logarithmic function is the inverse of an exponential function. If you start with

f(x) = b^x

then its inverse is

f^{-1}(x) = \log_b(x)

Thinking “inverse” helps you predict key features without memorizing them:

• Exponentials take any real input and output positive numbers.
• Logs take positive inputs and output any real number.

So the domain and range swap:

• For y = b^x, domain is all real numbers and range is y > 0.
• For y = \log_b(x), domain is x > 0 and range is all real numbers.

How base affects the shape (increasing vs decreasing)

The base b controls whether the log function increases or decreases.

• If

b > 1

then y = \log_b(x) is increasing.

• If

0 < b < 1

then y = \log_b(x) is decreasing.

This matches the behavior of exponentials: when 0 < b < 1, exponential functions decrease, and their inverses (logs) also reflect that reversal in a consistent way.

Key anchor points you should know

These come directly from the inverse relationship.

• Since

b^0 = 1

it follows that

\log_b(1) = 0

• Since

b^1 = b

it follows that

\log_b(b) = 1

Also, for any valid base,

\log_b(b^x) = x

and

b^{\log_b(x)} = x

These two “undoing” statements are the heart of simplifying and solving.

Worked examples

Example 1: Convert between logarithmic and exponential form

Rewrite

\log_5(125) = 3

in exponential form.

Because \log_b(x) = y means b^y = x, we get:

5^3 = 125

Example 2: Evaluate a log using the meaning

Evaluate

\log_2\left(\frac{1}{8}\right)

You are asking: what exponent on 2 gives \frac{1}{8}? Since

2^{-3} = \frac{1}{8}

it follows that

\log_2\left(\frac{1}{8}\right) = -3

Common misconceptions to avoid (built into the definition)

A frequent error is trying to treat \log_b(x) as if it were “something you can distribute over addition” or as if it were a product. But the log is not a multiplication symbol—it is a function. For example,

\log_b(3 + 7) \ne \log_b(3) + \log_b(7)

There are specific rules (properties) for products and quotients, not for sums.

Exam Focus

• Typical question patterns:
  • Convert between logarithmic and exponential form and interpret what a logarithm “means.”
  • Identify domain/range of y = \log_b(x) and evaluate logs using known exponent facts.
  • Use inverse relationships to simplify expressions like \log_b(b^x).
• Common mistakes:
  • Forgetting the input restriction x > 0 and allowing zero/negative arguments.
  • Using an invalid base (like b = 1 or b < 0).
  • Treating \log_b(x) like multiplication (for example, “canceling” parts that do not follow log rules).

Properties of Logarithms

Why properties exist (connect back to exponent rules)

Log properties are not random tricks; they come directly from exponent properties. Because logs are inverses of exponentials, any rule about multiplying powers becomes a rule about adding logs, and so on.

Start from the exponent facts:

b^m b^n = b^{m+n}

\frac{b^m}{b^n} = b^{m-n}

(b^m)^p = b^{mp}

Now translate each statement into “log language” by using the definition \log_b(x) = y \Leftrightarrow b^y = x.

Product, quotient, and power rules

If b > 0 and b \ne 1 and the arguments are positive, then:

Product Rule

\log_b(MN) = \log_b(M) + \log_b(N)

Quotient Rule

\log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N)

Power Rule

\log_b\left(M^p\right) = p\log_b(M)

The restrictions matter: you need M > 0 and N > 0 for the expressions to stay in the real numbers.

Why “no sum rule” is a big deal

There is no valid identity that turns \log_b(M + N) into something simpler in general. That means if you see a log of a sum, your options are limited: you might factor inside the log, or you might have to use numerical methods/approximations (depending on the problem), but you cannot “split” it the way you can with products.

Change of base (how to compute logs with any base)

In practice, calculators typically have buttons for \log(x) (base 10) and \ln(x) (base e). To evaluate \log_b(x) for another base, you use the change of base formula:

\log_b(x) = \frac{\log(x)}{\log(b)}

or

\log_b(x) = \frac{\ln(x)}{\ln(b)}

Both are correct; choose whichever matches your calculator tools.

Why it works (conceptually): you are converting “exponents in base b” into “exponents in base 10” or “exponents in base e.” Since all logarithms are just “exponent questions,” you can express one in terms of another.

Simplifying and expanding logs (and why each is useful)

You’ll often be asked to either:

• Expand a single logarithm into a sum/difference (using product/quotient/power rules). This is useful when matching forms, comparing expressions, or preparing to differentiate later (in calculus).
• Condense multiple logs into one (reverse the rules). This is useful for solving equations because one log equals another log is a common setup.

Worked examples

Example 1: Expand

Expand:

\log_3\left(\frac{9x^2\sqrt{y}}{5}\right)

Start by splitting quotient into subtraction:

\log_3\left(\frac{9x^2\sqrt{y}}{5}\right) = \log_3(9x^2\sqrt{y}) - \log_3(5)

Split the product into a sum:

\log_3(9x^2\sqrt{y}) = \log_3(9) + \log_3(x^2) + \log_3(\sqrt{y})

Now apply the power rule. Note that \sqrt{y} = y^{1/2}:

\log_3(x^2) = 2\log_3(x)

\log_3(\sqrt{y}) = \log_3\left(y^{1/2}\right) = \frac{1}{2}\log_3(y)

So the full expansion is:

\log_3(9) + 2\log_3(x) + \frac{1}{2}\log_3(y) - \log_3(5)

Example 2: Condense

Condense into a single logarithm:

2\ln(x) - \frac{1}{3}\ln(y) + \ln(5)

Use the power rule in reverse:

2\ln(x) = \ln\left(x^2\right)

\frac{1}{3}\ln(y) = \ln\left(y^{1/3}\right)

Now combine using product/quotient rules:

\ln\left(x^2\right) - \ln\left(y^{1/3}\right) + \ln(5) = \ln\left(\frac{5x^2}{y^{1/3}}\right)

Common pitfalls with properties

A very common mistake is misusing the power rule, especially when negatives are involved. For instance, students sometimes write

\log_b(x - 3)^2 = 2\log_b(x - 3)

This is ambiguous because the exponent might be outside the log’s argument depending on parentheses. The safe version is:

• If you mean \log_b\left((x - 3)^2\right), then you can use the power rule.
• If you mean \left(\log_b(x - 3)\right)^2, that is completely different and you cannot simplify it with log properties.

Parentheses are not optional here.

Exam Focus

• Typical question patterns:
  • Expand or condense logarithmic expressions using product/quotient/power rules.
  • Use change of base to evaluate a log with a nonstandard base or to compare expressions.
  • Justify an algebraic step in a simplification (showing you know which rule applies).
• Common mistakes:
  • Trying to “split” \log_b(M + N) or combine it as if a sum were a product.
  • Dropping necessary parentheses and changing the meaning of the expression.
  • Forgetting domain restrictions when rewriting (for example, condensing into a single log with an argument that must stay positive).

Solving Exponential and Logarithmic Equations

The big strategy: isolate the exponential or the logarithm

Solving these equations is mostly about choosing the right “undo” operation.

• If the variable is in the exponent (exponential equation), you typically take a logarithm of both sides.
• If the variable is inside a logarithm (logarithmic equation), you typically rewrite in exponential form or condense logs and then exponentiate.

You should also think about checking solutions, because logs can introduce restrictions. If you ever take a step that assumes the argument is positive, you must reject any final answers that violate that.

Solving exponential equations

Case 1: Same base on both sides

If you can rewrite both sides with the same base, then equate exponents. For example, if

3^{2x - 1} = 3^{x + 5}

then (because the exponential function with base 3 > 1 is one-to-one), you can set exponents equal:

2x - 1 = x + 5

and solve:

x = 6

Case 2: Different bases or messy expressions (use logs)

If you have something like

5^{x} = 17

there is no simple way to rewrite 17 as a power of 5, so you take logs:

\log\left(5^{x}\right) = \log(17)

Now apply the power rule:

x\log(5) = \log(17)

So

x = \frac{\log(17)}{\log(5)}

You could also use natural logs:

x = \frac{\ln(17)}{\ln(5)}

Both are correct.

Solving logarithmic equations

Case 1: One logarithm equals a number

Example structure:

\log_2(x - 1) = 5

Rewrite in exponential form:

2^5 = x - 1

So

32 = x - 1

x = 33

Now check the domain restriction for the original log:

x - 1 > 0

x > 1

The solution 33 is valid.

Case 2: Log equals log (same base)

If

\log_b(A) = \log_b(C)

then, because a log function is one-to-one,

A = C

as long as both arguments are in the domain (positive). Often you first condense each side into a single log.

Extraneous solutions (why checking matters)

When you solve log equations, you might create solutions that make a log’s argument nonpositive. Those are not allowed and must be thrown out.

A classic example:

\ln(x - 2) = \ln(1 - x)

If you use the “log equals log” rule, you would set:

x - 2 = 1 - x

So

2x = 3

x = \frac{3}{2}

Now check the original domain requirements:

x - 2 > 0 \Rightarrow x > 2

1 - x > 0 \Rightarrow x < 1

No real number can satisfy both, so the equation has no real solution. The algebra produced a number, but the original logarithms were never defined there.

Worked examples

Example 1: Exponential equation requiring logs

Solve:

2^{x+1} = 15

Take logs (base 10 or e are both fine). Using natural logs:

\ln\left(2^{x+1}\right) = \ln(15)

Power rule:

(x+1)\ln(2) = \ln(15)

Solve for x:

x+1 = \frac{\ln(15)}{\ln(2)}

x = \frac{\ln(15)}{\ln(2)} - 1

Example 2: Log equation with multiple terms

Solve:

\log_3(x) + \log_3(x - 2) = 2

First, note domain restrictions:

x > 0

x - 2 > 0 \Rightarrow x > 2

Now condense using the product rule:

\log_3\left(x(x - 2)\right) = 2

Rewrite in exponential form:

3^2 = x(x - 2)

9 = x^2 - 2x

x^2 - 2x - 9 = 0

Solve the quadratic:

x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-9)}}{2}

x = \frac{2 \pm \sqrt{4 + 36}}{2}

x = \frac{2 \pm \sqrt{40}}{2}

x = 1 \pm \sqrt{10}

Now enforce x > 2:

• 1 + \sqrt{10} is greater than 2, so it’s valid.
• 1 - \sqrt{10} is negative, so it violates x > 0 (and x > 2).

Final solution:

x = 1 + \sqrt{10}

A helpful “decision rule”

If you feel stuck, ask: where is the variable?

• Variable in an exponent → apply a logarithm.
• Variable inside a log → rewrite to exponential form after isolating/condensing logs.

That one question often chooses the method for you.

Exam Focus

• Typical question patterns:
  • Solve exponential equations by rewriting with a common base or by taking logs on both sides.
  • Solve log equations by condensing to one log, converting to exponential form, and checking for extraneous solutions.
  • Interpret solutions in context (for example, time in a growth model).
• Common mistakes:
  • Forgetting domain restrictions and reporting extraneous solutions.
  • Using log properties incorrectly (especially condensing/expanding errors).
  • Taking logs of a negative expression during a step (which is invalid in real numbers).

Logarithmic Function Graphs and Transformations

Start from the parent function

The “parent” logarithmic function is

y = \log_b(x)

Its most important graphical features come from being the inverse of

y = b^x

Core features you should be able to state quickly

For b > 1 (the common increasing case):

• Domain:

x > 0

• Range: all real numbers
• Vertical asymptote:

x = 0

• Intercept: passes through

(1,0)

because \log_b(1) = 0.

For 0 < b < 1, the same domain/asymptote hold, but the graph is decreasing.

Why the asymptote happens (conceptual)

The vertical asymptote at x = 0 isn’t a random rule. As x approaches 0 from the positive side, you are asking for the exponent that produces a number extremely close to zero.

• If b > 1, then very negative exponents make b^y tiny, so \log_b(x) heads toward negative infinity as x approaches 0^+.
• That’s why the graph dives downward near the y-axis but never actually touches x = 0.

Transformations of logarithmic graphs

A general transformed log function often looks like:

y = a\log_b(x - h) + k

Each parameter changes the graph in a predictable way.

• h shifts the graph horizontally. The vertical asymptote moves from x = 0 to

x = h

and the domain becomes

x > h

• k shifts the graph vertically.
• a stretches/compresses vertically; if a < 0, it reflects across the x-axis.

A common mistake is to think the asymptote moves to x = -h, but because the function uses x - h, the asymptote is where x - h = 0, which is x = h.

Transformations with a coefficient inside (horizontal scaling)

If you have

y = \log_b(c(x - h)) + k

the inside factor c affects horizontal scaling (and may also affect the domain depending on the sign). A safe way to reason is:

1. The asymptote is still where the argument is zero:

c(x - h) = 0 \Rightarrow x = h

2. The domain is where the argument is positive:

c(x - h) > 0

If c > 0, then this reduces to x > h. If c < 0, then the inequality flips and you get x < h.

Finding intercepts and key points (how to do it systematically)

To find an x-intercept, set y = 0 and solve.

To find a y-intercept, you would plug in x = 0—but for logarithms, x = 0 is usually not in the domain (unless the function is shifted). So many log graphs have **no** y-intercept.

A good “anchor point” method: because the parent passes through (1,0), the transformed graph of

y = a\log_b(x - h) + k

passes through the point where x - h = 1, meaning x = h + 1. Then

y = a\log_b(1) + k = k

So a reliable point is:

(h + 1, k)

This is often faster and less error-prone than picking random inputs.

Worked examples

Example 1: Describe features of a transformed log

Given

f(x) = \log_2(x - 3) + 1

• Vertical asymptote occurs when x - 3 = 0, so:

x = 3

• Domain is:

x > 3

• Range is all real numbers (vertical shifts do not change the range for logs).
• A key point comes from x - 3 = 1, so x = 4. Then:

f(4) = \log_2(1) + 1 = 1

So the graph passes through (4,1).

Example 2: Solve for an intercept and interpret domain

For

g(x) = -2\ln(x + 1) + 3

• Domain comes from x + 1 > 0:

x > -1

So the vertical asymptote is:

x = -1

• To find the x-intercept, set g(x) = 0:

-2\ln(x + 1) + 3 = 0

-2\ln(x + 1) = -3

\ln(x + 1) = \frac{3}{2}

Convert to exponential form (base e):

e^{3/2} = x + 1

x = e^{3/2} - 1

This intercept is valid because e^{3/2} - 1 > -1, so it lies in the domain.

Connecting graphs to real contexts

Logs appear when you measure something on a “multiplicative” scale but want an “additive” comparison.

• If one quantity is multiplied by a factor, a logarithm turns that multiplication into addition (via the product rule). That’s why logarithmic scales are used to compare very large ranges.
• In growth/decay problems, you often solve for time. Time ends up in the exponent, and taking logs is how you isolate it. Graphically, that corresponds to the inverse relationship between exponentials and logs.

Exam Focus

• Typical question patterns:
  • Identify domain, range, and vertical asymptote for transformed log functions.
  • Describe transformations from y = \log_b(x) to a given function and sketch using key points.
  • Find intercepts or specific points on a logarithmic graph using algebra and log definitions.
• Common mistakes:
  • Misplacing the vertical asymptote (confusing x = h with x = -h).
  • Claiming a y-intercept exists without checking whether x = 0 is in the domain.
  • Forgetting that the argument must stay positive after transformations (especially when there is a negative coefficient inside the log).