Exponential Growth and Decay

Ever wonder how your money grows in a bank or how populations increase over time? That's exponential growth in action! The basic form is y = a·b^x, where 'a' is your starting value and 'b' is the growth factor.

For practical applications, we often use y = a$1+r$^x, where 'r' is the rate of growth or decay (negative for decay). When dealing with money, we use the compound interest formula: A = P$1+r/n$^nt, where P is principal, r is rate, n is compounding frequency, and t is time.

Half-life problems (like radioactive decay) use the formula y = a(0.5)^x, where x represents the number of half-lives that have passed.

💡 How to tell if a relationship is exponential: Look at the pattern of change! If values are being multiplied by the same factor each time (like 2, 4, 8, 16...), it's exponential. If they increase by the same amount (like 2, 4, 6, 8...), it's linear.

When graphing exponential functions, remember they're transformations of parent functions like y = b^x. The domain is all real numbers, and the range is y > 0, with the x-axis $y = 0$ as an asymptote.

Graphing Exponential Functions

Exponential functions with different bases create distinct graphs that you can easily sketch with a few key points. For functions like y = (1/5)^x or y = 3^x, creating a table of values helps visualize the shape.

When x = 0, y always equals 1 for basic exponential functions $b^0 = 1$. For negative x-values, you're finding reciprocals $like 3^-2 = 1/9$. This creates the characteristic curve that approaches but never touches the x-axis.

When working with transformations like y = -2(3)^$x-1$, break it down step by step:

• The negative sign flips the graph across the x-axis
• The 2 stretches the graph vertically
• The $x-1$ shifts the graph right by 1 unit

🔑 Remember: All exponential functions have a horizontal asymptote. For basic functions like y = b^x, this asymptote is y = 0 $the x-axis$.

Understanding these transformations gives you the power to predict how any exponential function will behave without plotting every point.

Solving Exponential Equations

Exponential equations might look intimidating, but they follow a simple principle: if b^x = b^y, then x = y. This means the exponents must be equal when the bases are the same.

To solve exponential equations:

1. Make sure both sides have the same base
2. If they don't, rewrite one or both sides to get matching bases
3. Once the bases match, set the exponents equal

For example, with 3^x = 9^4, rewrite 9^4 as $3^2$^4 = 3^8, so x = 8.

Sometimes you need to isolate the exponential term first. For 53^-2x = 625, rewrite 625 as 5^4, then set 3-2x = 4, and solve for x to get x = 3/2.

💪 You can create an exponential model when given points! If you know (0,13210) and (7,15129), plug these into y = ab^x and solve for the unknown values. This skill is essential for real-world applications.

Remember that exponential growth and decay models are powerful tools for making predictions about everything from population growth to radioactive decay.

Introduction to Logarithms

Logarithms are basically the inverse of exponential functions - they help us solve for the exponent! If b^x = y, then log_b(y) = x. Think of logarithms as asking, "To what power must I raise the base to get this number?"

For example, log_3(9) = 2 because 3^2 = 9. Similarly, log_10(100) = 2 because 10^2 = 100, and log_10(0.01) = -2 because 10^-2 = 0.01.

When evaluating logarithms, remember you can rewrite them in exponential form. To find log_5(243), think: 5^? = 243. Since 5^5 = 3125 and 5^4 = 625, we need to solve this algebraically to get the exact answer.

🧠 Logarithms and exponentials are two sides of the same coin! Whenever you're stuck with a logarithm, try rewriting it as an exponential equation.

Graphing logarithmic functions follows similar transformation rules to other functions. For y = log_3(x), create a table using key values, then apply any transformations like y = log_3$x+2$+1. Remember that logarithmic functions have a vertical asymptote at x = 0, and their domain is x > 0.

Solving Logarithmic Equations

Logarithmic equations can be solved using two main strategies, depending on their structure:

1. For equations with one logarithm: Isolate the log, then convert to exponential form and solve. Example: log_8(x) = 4/3, convert to 8^(4/3) = x, so x = 16.

2. For equations with multiple logs of the same base: Use the property of equality - if the logs are equal, their arguments must be equal. Example: log_4(x²) = log_4$-6x-8$ means x² = -6x-8.

Always check your solutions! Remember that logarithms of negative numbers or zero are undefined, so potential solutions that give these values must be rejected.

⚠️ The most common mistake in logarithmic equations is forgetting to check your answers. Since logs have domain restrictions, not all algebraic solutions are valid!

When solving an equation like log_2$x²-18$ = log_2$-3x$, set x²-18 = -3x and solve the quadratic x²+3x-18 = 0 to get x = -6 or x = 3. But check both: log_2(-18-18) would involve a negative number inside a log, which is invalid, so x = 3 is your only solution.

Properties of Logarithms

Logarithm properties are your secret weapons for simplifying complex expressions! These three key properties make working with logs much easier:

1. Product Rule: log_b(m·n) = log_b(m) + log_b(n)
2. Quotient Rule: log_b$m/n$ = log_b(m) - log_b(n)
3. Power Rule: log_b$m^n$ = n·log_b(m)

To expand logarithmic expressions, break them apart using these rules. For instance, log_2(4x²y) becomes log_2(4) + log_2(x²) + log_2(y), which simplifies to 2 + 2log_2(x) + log_2(y).

Conversely, to write multiple logarithms as a single logarithm, work backwards. For log_2(6) + log_2(3) - log_2(15), combine them as log_2(6·3/15) = log_2(18/15) = log_2(6/5).

💡 A useful trick: log_b(1) = 0 for any base b, because b^0 = 1. This helps simplify many expressions!

These properties also help solve logarithmic equations. For log_3(x) + log_3(6) = 4, combine the logs to get log_3(6x) = 4, then convert to exponential form: 3^4 = 6x, so x = 81/6 = 13.5.

Advanced Logarithmic Problem Solving

What happens when you need to solve exponential equations with different bases? This is where logarithms really shine! The change-of-base formula lets you convert between any logarithmic bases: log_b(a) = log(a)/log(b).

To evaluate log_5(140), use the formula: log_5(140) = log(140)/log(5) ≈ 3.0704.

When solving exponential equations like 5^x = 62:

1. Take the logarithm of both sides (any base works, but natural log or log base 10 is common)
2. Use logarithm properties to isolate x
3. Solve for x using the change-of-base formula

For example, with 5^x = 62:

• log_5(62) = x
• log(62)/log(5) = x
• x ≈ 2.5643

🧮 When solving exponential equations where you can't easily match bases, logarithms are your best tool! This technique works for any exponential equation.

For more complex equations like 3^(2x)+5 = 62, first isolate the exponential term $3^(2x) = 57$, then take logarithms and solve: x = log(57)/(2log(3)) ≈ 1.8401.

Natural Logarithms and the Number e

The number e ≈ 2.71828 is one of the most important constants in mathematics. It appears naturally in growth and decay problems, which is why logarithms with base e are called "natural logarithms" (written as ln).

Natural logarithms work just like other logarithms, but they're especially useful for continuous growth models. Instead of using A = P$1+r/n$^nt for compound interest, we can use A = Pe^(rt) for continuous compounding.

To solve equations with e or natural logs:

• For e^x = 23, take the natural log of both sides: ln$e^x$ = ln(23), which gives x = ln(23)
• For ln(25) = x, convert to exponential form: e^x = 25

🌱 The number e appears in nature everywhere, from population growth to radioactive decay. That's why natural logarithms are so valuable in science and finance!

Properties of logarithms work the same with natural logs. For example, 2ln(3) + ln(4) + ln(y) = ln$3^2 · 4 · y$ = ln(36y).

When solving real-world problems, like finding how long it takes $700 to grow to $800 at 3% interest compounded continuously, set up 800 = 700e^(0.03t) and solve for t: t = ln(800/700)/0.03 ≈ 4.7 years.

Adding and Subtracting Rational Expressions

Adding and subtracting rational expressions is like working with fractions - you need a common denominator. Here's your game plan:

1. Factor all denominators completely
2. Find the Least Common Denominator (LCD) - the product of all unique factors, each raised to its highest power
3. Rewrite each fraction with the LCD as denominator
4. Add or subtract the numerators while keeping the common denominator

For example, to find the LCD of 15ab^3c, 16b^5c^2, and 20a^5c^6, identify the highest power of each variable needed: a^5, b^5, c^6, and relevant coefficients to get 240a^5b^5c^6.

When adding fractions like $5a^2/6b$ + $9/7a^4b^3$, first find the LCD: 42a^4b^3. Then convert each fraction to equivalent fractions with this denominator and add the numerators.

🔑 Always factor denominators before finding the LCD! This makes the process much more efficient.

For more complex expressions like $x+10$/$3x-15$ - $3x+15$/$6x-30$, factor the denominators first: 3$x-5$ and 6$x-5$. Then use the LCD of 6$x-5$ to combine them, resulting in the simplified answer of -1/6.

Solving Rational Equations

Rational equations involve fractions with variables, and they require special care to solve correctly. Follow these steps:

1. If one fraction on each side: Cross multiply to clear denominators
2. If multiple fractions: Multiply every term by the LCD to clear all denominators
3. Solve the resulting equation (usually polynomial)
4. Always check your answers by substituting back into the original equation

For example, to solve $x-3$/$2x+3$ = $x-5$/$x-1$:

1. Cross multiply: $x-3$$x-1$ = $x-5$$2x+3$
2. Expand: x²-4x+3 = 2x²-5x-15
3. Rearrange to standard form: -x²+x+18 = 0 or x²-x-18 = 0
4. Factor and solve: $x-6$$x+3$ = 0, so x = 6 or x = -3

⚠️ Extraneous solutions can sneak in when solving rational equations. Always check your answers in the original equation to make sure they don't cause division by zero!

For equations like -12/x + 24/$x-2$ = 4, multiply all terms by the LCD x$x-2$ to get -12$x-2$ + 24x = 4x$x-2$. Then solve the resulting quadratic equation to find that x = 6 or x = -2.