Exponentials
Exponentials, expressed as ab, where 'a' is the base and 'b' is the exponent, play a crucial role in representing exponential growth or decay, compound interest, and numerous other mathematical and real-world applications.
Properties of Exponentials
- Product of Powers: am * an = a(m+n)When multiplying two exponential terms with the same base, add the exponents.
- Quotient of Powers: am / an = a(m-n)When dividing two exponential terms with the same base, subtract the exponents.
- Power of a Power: (am)n = a(mn)When an exponential term is raised to another power, multiply the exponents.
- Power of a Product: (ab)m = am * bmThe exponent applies to each base when a product is raised to a power.
- Negative Exponent: a(-m) = 1/amA negative exponent indicates the reciprocal of the base raised to the absolute value of the exponent.
- Zero Exponent: a0 = 1 (where a ≠ 0)Any non-zero number raised to the power of zero is 1.
Applications of Exponentials
Exponential functions are pivotal in real-world scenarios like population growth, radioactive decay, and financial calculations involving compound interest.
Example Question 1: Exponential Growth
A population of bacteria doubles every 3 hours. If initially, there are 50 bacteria, how many will there be after 12 hours?
Solution:
Using the formula for exponential growth: P(t) = P0 * 2(t/T), where P0 is the initial population, T is the doubling time, and t is the time elapsed.
P(12) = 50 * 2(12/3) = 50 * 24 = 50 * 16 = 800
After 12 hours, there will be 800 bacteria.
Logarithms
Logarithms, the inverses of exponential functions, are expressed as log_b(a), indicating the power to which 'b' must be raised to obtain 'a'. They are fundamental in solving exponential equations and find applications in various scientific calculations.
Properties of Logarithms
- Product: logb(mn) = logb(m) + logb(n)The logarithm of a product is the sum of the logarithms of the factors.
- Quotient: logb(m/n) = logb(m) - logb(n)The logarithm of a quotient is the difference of the logarithms.
- Power: logb(mn) = n * logb(m)The logarithm of a power is the exponent times the logarithm of the base.
- Change of Base Formula: logb(a) = logc(a) / logc(b)The base of a logarithm can be changed using this formula.
Applications of Logarithms
Logarithms are instrumental in solving problems related to exponential decay, pH chemistry, and Richter scale measurements, among others.
Example Question 2: Solving Exponential Equations
Solve for x in the equation 3(2x+1) = 81.
Solution:
Taking the logarithm on both sides: log(3(2x+1)) = log(81)
Using the property of logarithms: (2x + 1) * log(3) = log(81)
Since 81 is 34: (2x + 1) * log(3) = 4 * log(3)
Solving for x: 2x + 1 = 4 => x = 1.5
Thus, x = 1.5 is the solution to the equation.
FAQ
The change of base formula allows you to compute the logarithm of a number in one base using logarithms in another base. The formula is logb(a) = logc(a) / logc(b), where c can be any positive number except 1. Essentially, you are taking the logarithm of 'a' using base 'c' and dividing it by the logarithm of 'b' using base 'c'. This formula is particularly useful when you have a calculator that only computes logarithms in base 10 or base e, but you need the logarithm in another base.
Negative exponents in exponential functions, like b(-a), can be rewritten as 1 / ba. The negative exponent indicates that the base should be taken as the reciprocal and then raised to the positive of the exponent. For example, 2(-3) is equivalent to 1 / 23 = 1/8. This property allows us to simplify expressions and solve equations that involve negative exponents, ensuring that we work with positive exponents, which are often easier to handle and understand.
The number 'e', approximately equal to 2.71828, is the base of the natural logarithm and has unique properties when used in exponential functions. It is the only function where the derivative is equal to the original function, making it particularly useful in calculus, particularly in problems involving continuous compound interest, population growth, or decay problems. The function f(x) = ex has a slope of ex at any point, which is not true for other bases. This property simplifies the analysis of functions and differential equations involving 'e'.
In mathematics, a logarithm is defined only for positive real numbers. Specifically, if you have a logarithm in the form logb(a), 'a' must be greater than zero (a > 0) for the logarithm to be defined. If 'a' is less than or equal to zero, the logarithm is undefined. This is because logarithms represent the power to which a base must be raised to produce a number, and there is no exponent to which you can raise a positive number to get a non-positive number. Always check the domain of the function to ensure the logarithm is defined.
Yes, logarithms are particularly useful in solving exponential equations, even those with different bases. By applying logarithmic properties, you can simplify and solve equations with different bases. For example, to solve an equation like 2x = 5, you can take the logarithm of both sides of the equation, choosing a convenient logarithm base, often e (natural logarithm) or 10. Using log properties, you can bring down the exponent as a multiplier: x * log(2) = log(5), and solve for x by dividing by log(2).
Practice Questions
To solve the equation 5(2x + 3) = 125(x - 1), we must express both sides of the equation with the same base. Recognising that 125 is equal to 53, we rewrite the equation as 5(2x + 3) = (53)(x - 1). Using the property of exponents that (am)n = a(mn), we simplify to get 5(2x + 3) = 5(3x - 3). Since the bases are the same, we can set the exponents equal to each other: 2x + 3 = 3x - 3. Solving for x, we find that x = 6. Therefore, the solution to the equation 5(2x + 3) = 125(x - 1) is x = 6.
To solve the logarithmic equation log_3(x + 5) + log3(x - 2) = 2, we can use the logarithm property that logb(m) + logb(n) = logb(mn) to combine the two logarithms on the left-hand side of the equation: log3((x + 5)(x - 2)) = 2. Next, we can rewrite the equation in exponential form to eliminate the logarithm: (x + 5)(x - 2) = 32. Simplifying, we get x2 + 3x - 10 = 9. Moving the 9 to the left-hand side of the equation to set it equal to zero, we get x2 + 3x - 19 = 0. To find the roots of x2 + 3x - 19 = 0, we can use the quadratic formula. Plugging in the coefficients A = 1, B = 3, and C = -19, we get two potential solutions for x: x = (-3 + sqrt(85)) / 2 and x = (-3 - sqrt(85)) / 2. It's crucial to check these solutions back in the original logarithmic equation to ensure they don't cause the argument of the logarithm to be non-positive, as the logarithm is only defined for positive numbers.
