6.3 Logarithmic Functions

Logarithmic Functions

A logarithm answers the question: "What exponent do I need?" If you know that $2^3 = 8$, then $\log_2(8) = 3$ just asks, "What power of 2 gives me 8?" Logarithmic functions are the inverses of exponential functions, and that inverse relationship is the key to everything in this section.

Logarithmic and Exponential Form Conversion

The core relationship to memorize is:

$\log_b(x) = y \quad \text{is equivalent to} \quad b^y = x$

• $b$ is the base (must be positive and not equal to 1)
• $x$ is the argument (the number you're taking the log of)
• $y$ is the exponent/logarithm (the answer)

Converting logarithmic → exponential form: Take the base of the log, raise it to the result on the other side of the equation, and set it equal to the argument.

• $\log_2(8) = 3$ becomes $2^3 = 8$
• $\log_5(25) = 2$ becomes $5^2 = 25$

Converting exponential → logarithmic form: Identify the base, the exponent, and the result, then rearrange.

• $10^4 = 10000$ becomes $\log_{10}(10000) = 4$
• $3^{-2} = \frac{1}{9}$ becomes $\log_3\left(\frac{1}{9}\right) = -2$

This conversion works because logarithmic and exponential functions are inverses of each other. If $f(x) = b^x$, then $f^{-1}(x) = \log_b(x)$.

Evaluation of Logarithms with Bases

Two bases show up constantly:

• Common logarithm: base 10, written as $\log(x)$ or $\log_{10}(x)$
• Natural logarithm: base $e$, written as $\ln(x)$ or $\log_e(x)$, where $e \approx 2.71828$

Your calculator has buttons for both of these, but what if you need $\log_3(20)$? That's where the change of base formula comes in:

$\log_b(x) = \frac{\log_a(x)}{\log_a(b)}$

You can pick any base $a$, but base 10 or base $e$ are the practical choices since your calculator handles them. For example:

$\log_3(20) = \frac{\log(20)}{\log(3)} = \frac{1.3010}{0.4771} \approx 2.727$

One restriction to remember: logarithms can only be evaluated for positive arguments ($x > 0$). You cannot take the log of zero or a negative number.

Domain, Range, and Asymptotes of Logarithmic Functions

For the parent function $f(x) = \log_b(x)$:

• Domain: all positive real numbers ($x > 0$)
• Range: all real numbers ($-\infty < y < \infty$)
• Vertical asymptote: $x = 0$

As $x$ approaches 0 from the right, the function drops toward $-\infty$. As $x$ increases, the function grows, but more and more slowly. The graph always passes through the point $(1, 0)$ because $\log_b(1) = 0$ for any valid base, and through $(b, 1)$ because $\log_b(b) = 1$.

If the function is transformed, like $f(x) = \log_b(x - h) + k$, the domain shifts to $x > h$ and the vertical asymptote moves to $x = h$.

Applications of Logarithms

Common Logarithms in Real-World Applications

Logarithmic scales are used when data spans a huge range of values. Instead of dealing with numbers from 0.0000001 to 10,000,000, a log scale compresses that into a manageable range.

Richter Scale (earthquake magnitude):

$M = \log\left(\frac{\text{Seismic wave amplitude}}{\text{Standard amplitude}}\right)$

Because this is a log base 10 scale, an increase of 1 on the Richter scale means the seismic wave amplitude is 10 times larger. A magnitude 6 earthquake has seismic waves 10 times the amplitude of a magnitude 5.

Decibel Scale (sound intensity):

$\text{dB} = 10 \log\left(\frac{\text{Sound intensity}}{\text{Reference intensity}}\right)$

An increase of 10 dB corresponds to a tenfold increase in sound intensity. Normal conversation is about 60 dB; a rock concert at 110 dB is actually 100,000 times more intense.

pH Scale (acidity):

$\text{pH} = -\log([\text{H}^+])$

Here, $[\text{H}^+]$ is the hydrogen ion concentration in moles per liter. The negative sign means that as hydrogen ion concentration increases, pH decreases. A decrease of 1 pH unit means a tenfold increase in hydrogen ion concentration. Pure water has a pH of 7; lemon juice at pH 2 is 100,000 times more acidic.

Natural Logarithms for Growth and Decay

Exponential growth is modeled by $A(t) = A_0 e^{kt}$, where $A_0$ is the initial amount, $k$ is the growth rate (positive), and $t$ is time.

• Doubling time (how long it takes the quantity to double): $t_d = \frac{\ln(2)}{k}$

Exponential decay is modeled by $A(t) = A_0 e^{-kt}$, where $k$ is the decay rate (positive, but the negative sign in the exponent makes the quantity shrink).

• Half-life (how long it takes the quantity to halve): $t_{1/2} = \frac{\ln(2)}{k}$

Notice that doubling time and half-life use the same formula. That's because $\ln(2)$ comes from setting the quantity equal to twice (or half) the starting amount and solving.

Carbon dating example: Carbon-14 has a half-life of 5,730 years, so $k = \frac{\ln(2)}{5730} \approx 0.000121$. If a fossil has 25% of its original carbon-14, you can solve $0.25 = e^{-0.000121t}$ by taking the natural log of both sides:

$\ln(0.25) = -0.000121t$ $t = \frac{\ln(0.25)}{-0.000121} \approx 11{,}460 \text{ years}$

Solving Equations with Logarithmic Properties

Three properties do most of the heavy lifting:

• 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: $\log_b(M^n) = n \cdot \log_b(M)$

Steps to solve a logarithmic equation:

1. Isolate the logarithmic expression(s) on one side of the equation.
2. Combine logs using the product, quotient, or power rules so you have a single logarithm.
3. Convert to exponential form to eliminate the log.
4. Solve the resulting equation algebraically.
5. Check your answer(s) by plugging back into the original equation. Any solution that makes the argument of a logarithm zero or negative must be rejected.

Example: Solve $\log_2(x) + \log_2(x - 2) = 3$

1. Combine using the product rule: $\log_2(x(x-2)) = 3$

2. Convert to exponential form: $x(x-2) = 2^3 = 8$

3. Expand and solve: $x^2 - 2x - 8 = 0$, which factors to $(x-4)(x+2) = 0$

4. Solutions: $x = 4$ or $x = -2$

5. Check: $x = -2$ fails because $\log_2(-2)$ is undefined. So $x = 4$ is the only valid solution.

That final check step is easy to forget, but skipping it is one of the most common mistakes on exams.