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Functions - Properties of Exponential and Logarithmic Functions

Grade 12IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Exponential Functions: An exponential function is defined as f(x)=axf(x) = a^x where a>0a > 0 and a≠1a \neq 1. Visually, the graph passes through (0,1)(0, 1) and has a horizontal asymptote at y=0y = 0. If a>1a > 1, the graph rises steeply to the right (exponential growth), while if 0<a<10 < a < 1, the graph falls toward the x-axis (exponential decay).

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Logarithmic Functions as Inverses: The logarithmic function f(x)=log⁡a(x)f(x) = \log_a(x) is the inverse of f(x)=axf(x) = a^x. Geometrically, the graph of a logarithm is a reflection of the exponential graph across the diagonal line y=xy = x. While exponential functions have a horizontal asymptote, logarithmic functions have a vertical asymptote at x=0x = 0 and an x-intercept at (1,0)(1, 0).

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Domain and Range Constraints: For the basic exponential function y=axy = a^x, the domain is all real numbers R\mathbb{R} and the range is y>0y > 0. Conversely, for the logarithmic function y=log⁡a(x)y = \log_a(x), the domain is restricted to x>0x > 0 (you cannot take the log of zero or a negative number), and the range is all real numbers R\mathbb{R}.

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The Natural Logarithm and ee: The number e≈2.718e \approx 2.718 is a fundamental constant in calculus. The natural logarithm is written as ln⁡(x)\ln(x), which is shorthand for log⁡e(x)\log_e(x). The functions f(x)=exf(x) = e^x and f(x)=ln⁡(x)f(x) = \ln(x) are inverses, meaning eln⁡(x)=xe^{\ln(x)} = x and ln⁡(ex)=x\ln(e^x) = x.

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Laws of Logarithms: Logarithms follow specific algebraic rules that simplify expressions. The Product Law states log⁡a(xy)=log⁡a(x)+log⁡a(y)\log_a(xy) = \log_a(x) + \log_a(y), the Quotient Law states log⁡a(xy)=log⁡a(x)−log⁡a(y)\log_a(\frac{x}{y}) = \log_a(x) - \log_a(y), and the Power Law states log⁡a(xk)=k⋅log⁡a(x)\log_a(x^k) = k \cdot \log_a(x). These laws are essential for solving equations where the variable is in the exponent.

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Transformations of Graphs: For y=p⋅aq(x−h)+ky = p \cdot a^{q(x-h)} + k or y=p⋅log⁡a(q(x−h))+ky = p \cdot \log_a(q(x-h)) + k, the parameter hh represents a horizontal shift, kk represents a vertical shift, pp is a vertical stretch/compression, and qq is a horizontal stretch/compression. In exponential functions, kk determines the horizontal asymptote y=ky = k. In logarithmic functions, hh determines the vertical asymptote x=hx = h.

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Change of Base: To evaluate a logarithm with a base not available on a standard calculator, the change of base formula is used: log⁡a(b)=log⁡c(b)log⁡c(a)\log_a(b) = \frac{\log_c(b)}{\log_c(a)}. Usually, cc is chosen to be 1010 or ee for calculation purposes.

📐Formulae

y=ax  ⟺  x=log⁡a(y)y = a^x \iff x = \log_a(y)

log⁡a(MN)=log⁡a(M)+log⁡a(N)\log_a(MN) = \log_a(M) + \log_a(N)

log⁡a(MN)=log⁡a(M)−log⁡a(N)\log_a(\frac{M}{N}) = \log_a(M) - \log_a(N)

log⁡a(Mk)=k⋅log⁡a(M)\log_a(M^k) = k \cdot \log_a(M)

log⁡a(a)=1\log_a(a) = 1 and log⁡a(1)=0\log_a(1) = 0

log⁡a(b)=ln⁡(b)ln⁡(a)=log⁡10(b)log⁡10(a)\log_a(b) = \frac{\ln(b)}{\ln(a)} = \frac{\log_{10}(b)}{\log_{10}(a)}

eln⁡(x)=xe^{\ln(x)} = x and ln⁡(ex)=x\ln(e^x) = x

💡Examples

Problem 1:

Solve the equation 5x+1=32x5^{x+1} = 3^{2x} for xx. Give your answer in terms of natural logarithms.

Solution:

Step 1: Take the natural log of both sides: ln⁡(5x+1)=ln⁡(32x)\ln(5^{x+1}) = \ln(3^{2x}) Step 2: Use the power law to bring exponents down: (x+1)ln⁡(5)=2xln⁡(3)(x+1)\ln(5) = 2x\ln(3) Step 3: Expand the left side: xln⁡(5)+ln⁡(5)=2xln⁡(3)x\ln(5) + \ln(5) = 2x\ln(3) Step 4: Rearrange to group xx terms: ln⁡(5)=2xln⁡(3)−xln⁡(5)\ln(5) = 2x\ln(3) - x\ln(5) Step 5: Factor out xx: ln⁡(5)=x(2ln⁡(3)−ln⁡(5))\ln(5) = x(2\ln(3) - \ln(5)) Step 6: Solve for xx: x=ln⁡(5)2ln⁡(3)−ln⁡(5)x = \frac{\ln(5)}{2\ln(3) - \ln(5)} Step 7: (Optional) Simplify the denominator: x=ln⁡(5)ln⁡(9)−ln⁡(5)=ln⁡(5)ln⁡(1.8)x = \frac{\ln(5)}{\ln(9) - \ln(5)} = \frac{\ln(5)}{\ln(1.8)}

Explanation:

This problem uses the Power Law of logarithms to move variables from the exponent to the base level, allowing us to solve using standard linear algebra techniques.

Problem 2:

Given the function f(x)=ln⁡(x−2)+3f(x) = \ln(x - 2) + 3, find the domain, the range, and the equation of the vertical asymptote.

Solution:

Step 1: To find the domain, set the argument of the log to be greater than zero: x−2>0  ⟹  x>2x - 2 > 0 \implies x > 2. Step 2: The vertical asymptote occurs where the argument is zero: x−2=0  ⟹  x=2x - 2 = 0 \implies x = 2. Step 3: Since the range of a basic logarithmic function is all real numbers and vertical shifts do not change this, the range is y∈Ry \in \mathbb{R}.

Explanation:

Logarithmic functions are only defined for positive arguments. The horizontal shift inside the natural log dictates both the domain restriction and the location of the vertical asymptote.