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Functions - Exponential and Logarithmic Functions

Grade 11IB

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

🔑Concepts

The Exponential Function f(x)=axf(x) = a^x: An exponential function is defined by a constant base a>0a > 0 and a1a \neq 1. Visually, the graph is a smooth curve that always stays above the x-axis, meaning the x-axis (y=0y = 0) serves as a horizontal asymptote. If a>1a > 1, the graph rises steeply to the right (exponential growth), whereas if 0<a<10 < a < 1, it falls toward the x-axis (exponential decay). All such functions pass through the y-intercept (0,1)(0, 1).

The Natural Exponential Function f(x)=exf(x) = e^x: The constant e2.71828e \approx 2.71828 is the unique base for which the slope of the tangent line to the curve at the point (0,1)(0, 1) is exactly 11. Its graph is a specific case of exponential growth and is the primary tool for modeling continuous growth or decay processes in science and finance.

Logarithmic Functions as Inverses: The function f(x)=loga(x)f(x) = \log_a(x) is defined as the inverse of the exponential function f(x)=axf(x) = a^x. Geometrically, this means the graph of a logarithm is the reflection of the exponential graph across the diagonal line y=xy = x. Unlike exponential functions, logarithmic functions have a vertical asymptote at x=0x = 0 (the y-axis) and pass through the x-intercept (1,0)(1, 0). The domain is restricted to (0,)(0, \infty).

Laws of Logarithms: These properties allow for the simplification of complex logarithmic and exponential expressions. The Product Law logb(xy)=logbx+logby\log_b(xy) = \log_b x + \log_b y turns multiplication into addition; the Quotient Law logb(xy)=logbxlogby\log_b(\frac{x}{y}) = \log_b x - \log_b y turns division into subtraction; and the Power Law logb(xk)=klogbx\log_b(x^k) = k \log_b x allows exponents to be brought down as coefficients.

The Natural Logarithm: The natural logarithm, written as ln(x)\ln(x), is simply a logarithm with base ee. It is the most common logarithm used in higher-level mathematics. An important visual and algebraic property is that elnx=xe^{\ln x} = x and ln(ex)=x\ln(e^x) = x, demonstrating that they 'undo' each other.

Transformations of Graphs: For the general form y=paxh+ky = p \cdot a^{x-h} + k or y=ploga(xh)+ky = p \cdot \log_a(x-h) + k, the constant kk shifts the graph vertically and moves the horizontal asymptote (y=ky=k), while hh shifts the graph horizontally and moves the vertical asymptote (x=hx=h). The multiplier pp causes a vertical stretch or compression and can reflect the graph across the x-axis if p<0p < 0.

Change of Base Formula: This concept allows us to convert a logarithm with any base into a ratio of logarithms with a common base (usually base ee or base 1010). This is represented by logab=logcblogca\log_a b = \frac{\log_c b}{\log_c a}. Visually, changing the base of a logarithmic function results in a vertical stretch or compression of the curve.

📐Formulae

y=ax    x=logayy = a^x \iff x = \log_a y

logb(mn)=logbm+logbn\log_b(mn) = \log_b m + \log_b n

logb(mn)=logbmlogbn\log_b(\frac{m}{n}) = \log_b m - \log_b n

logb(mk)=klogbm\log_b(m^k) = k \log_b m

logab=lnblna\log_a b = \frac{\ln b}{\ln a}

alogax=xa^{\log_a x} = x and loga(ax)=x\log_a(a^x) = x

lne=1\ln e = 1 and ln1=0\ln 1 = 0

💡Examples

Problem 1:

Solve the equation for xx: 4x+2=504^{x+2} = 50.

Solution:

  1. Take the natural logarithm of both sides: ln(4x+2)=ln(50)\ln(4^{x+2}) = \ln(50).
  2. Use the Power Law to move the exponent: (x+2)ln4=ln50(x+2) \ln 4 = \ln 50.
  3. Isolate the term with xx: x+2=ln50ln4x + 2 = \frac{\ln 50}{\ln 4}.
  4. Subtract 2 from both sides: x=ln50ln42x = \frac{\ln 50}{\ln 4} - 2.
  5. Calculate the approximate value: x2.822/1.38622.0362=0.822x \approx 2.822 / 1.386 - 2 \approx 2.036 - 2 = 0.822.

Explanation:

To solve an exponential equation where the bases cannot be easily matched, we apply logarithms to both sides to bring the variable down from the exponent, then use standard algebra to isolate the variable.

Problem 2:

Solve for xx: log3(x)+log3(x8)=2\log_3(x) + \log_3(x-8) = 2.

Solution:

  1. Use the Product Law to combine the logarithms: log3(x(x8))=2\log_3(x(x-8)) = 2.
  2. Convert the equation to exponential form: x(x8)=32x(x-8) = 3^2.
  3. Simplify the expression: x28x=9x^2 - 8x = 9.
  4. Form a quadratic equation: x28x9=0x^2 - 8x - 9 = 0.
  5. Factor the quadratic: (x9)(x+1)=0(x-9)(x+1) = 0.
  6. Identify potential solutions: x=9x = 9 or x=1x = -1.
  7. Check validity: Logs are only defined for positive values. log3(9)\log_3(9) is valid, but log3(1)\log_3(-1) is undefined. Therefore, x=9x = 9.

Explanation:

This problem requires condensing the logs into a single term before converting to exponential form. It is critical to check the final answers against the original domain, as log functions cannot accept non-positive arguments.