krit.club logo

Algebra - Logarithmic and Exponential Functions

Grade 12A Level

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

🔑Concepts

•

The exponential function y=axy = a^x (where a>1a > 1) represents growth, while y=a−xy = a^{-x} or y=(1a)xy = (\frac{1}{a})^x represents decay. These functions have a horizontal asymptote at y=0y = 0.

Graph of the exponential growth function y = 2^x passing through (0,1).
•

The logarithmic function y=log⁡a(x)y = \log_a(x) is the inverse of the exponential function. Its graph is a reflection of y=axy = a^x across the line y=xy = x. It has a vertical asymptote at x=0x = 0 and passes through (1,0)(1, 0).

Graph of the logarithmic function y = log2(x) passing through (1,0).
•

Natural logarithms use base ee (Euler's number ≈2.718\approx 2.718). The function y=exy = e^x and its inverse y=ln⁡(x)y = \ln(x) are central to calculus and model continuous growth processes.

Comparison of e^x and ln(x) reflected across y=x.
•

Logarithms convert multiplicative relationships into additive ones, allowing for the solution of equations where the variable is in the exponent by taking the log of both sides.

📐Formulae

log⁡a(xy)=log⁡ax+log⁡ay\log_a (xy) = \log_a x + \log_a y (Product Rule)

log⁡a(xy)=log⁡ax−log⁡ay\log_a (\frac{x}{y}) = \log_a x - \log_a y (Quotient Rule)

log⁡a(xk)=klog⁡ax\log_a (x^k) = k \log_a x (Power Rule)

log⁡aa=1\log_a a = 1 and log⁡a1=0\log_a 1 = 0

log⁡ab=log⁡cblog⁡ca\log_a b = \frac{\log_c b}{\log_c a} (Change of Base)

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

ax=b  ⟺  x=log⁡blog⁡aa^x = b \iff x = \frac{\log b}{\log a}

💡Examples

Problem 1:

Solve for xx: 52x−1=125^{2x-1} = 12. Give your answer to 3 decimal places.

Solution:

2x−1=log⁡512  ⟹  2x−1=ln⁡12ln⁡5  ⟹  2x=2.48491.6094+1  ⟹  2x=2.5439  ⟹  x≈1.2722x - 1 = \log_5 12 \implies 2x - 1 = \frac{\ln 12}{\ln 5} \implies 2x = \frac{2.4849}{1.6094} + 1 \implies 2x = 2.5439 \implies x \approx 1.272

Explanation:

To solve an exponential equation where the bases cannot be made the same, take the logarithm of both sides. Use the power rule to bring the exponent down, then isolate xx using algebraic manipulation.

Problem 2:

Simplify the expression: 2log⁡36−log⁡342\log_3 6 - \log_3 4.

Solution:

2log⁡36−log⁡34=log⁡3(62)−log⁡34=log⁡336−log⁡34=log⁡3(364)=log⁡39=22\log_3 6 - \log_3 4 = \log_3 (6^2) - \log_3 4 = \log_3 36 - \log_3 4 = \log_3 (\frac{36}{4}) = \log_3 9 = 2.

Explanation:

First, apply the Power Rule to move the coefficient into the exponent. Then, use the Quotient Rule for logarithms to combine the terms. Finally, evaluate the resulting logarithm.

Problem 3:

Solve the equation: log⁡2x+log⁡2(x−2)=3\log_2 x + \log_2 (x - 2) = 3.

Solution:

log⁡2[x(x−2)]=3  ⟹  x2−2x=23  ⟹  x2−2x−8=0  ⟹  (x−4)(x+2)=0\log_2 [x(x - 2)] = 3 \implies x^2 - 2x = 2^3 \implies x^2 - 2x - 8 = 0 \implies (x - 4)(x + 2) = 0. Thus, x=4x = 4 or x=−2x = -2. However, xx must be >2> 2 for the logs to be defined, so x=4x = 4.

Explanation:

Use the Product Rule to combine the logarithms into a single term. Convert the logarithmic equation into its equivalent exponential form (ay=xa^y = x). Solve the resulting quadratic equation and always check for extraneous solutions (logs of negative numbers are undefined).

Problem 4:

Sketch the graph of y=3x−2y = 3^x - 2 and identify the horizontal asymptote and the yy-intercept.

Graph of y = 3^x - 2 showing horizontal asymptote at y = -2.

Solution:

  1. The base function is y=3xy = 3^x, which passes through (0,1)(0,1).
  2. The transformation −2-2 shifts the graph downwards by 2 units.
  3. The original horizontal asymptote y=0y = 0 becomes y=−2y = -2.
  4. To find the yy-intercept, set x=0x = 0: y=30−2=1−2=−1y = 3^0 - 2 = 1 - 2 = -1.
  5. The yy-intercept is (0,−1)(0, -1).

Explanation:

Exponential functions of the form y=ax+ky = a^x + k have a horizontal asymptote at y=ky = k. The vertical shift affects all points and the asymptote equally.

Problem 5:

The power PP in a circuit is related to time tt by P=100e−0.5tP = 100 e^{-0.5t}. Find the value of tt when the power drops to 2020 units.

Decay curve of P = 100e^-0.5t showing the point where P = 20.

Solution:

  1. Set up the equation: 20=100e−0.5t20 = 100 e^{-0.5t}.
  2. Divide by 100: 0.2=e−0.5t0.2 = e^{-0.5t}.
  3. Take the natural logarithm of both sides: ln⁡(0.2)=−0.5t\ln(0.2) = -0.5t.
  4. Solve for tt: t=ln⁡(0.2)−0.5t = \frac{\ln(0.2)}{-0.5}.
  5. t≈−1.609−0.5≈3.218t \approx \frac{-1.609}{-0.5} \approx 3.218.

Explanation:

To solve for a variable in the exponent of ee, use the natural logarithm ln⁡\ln because ln⁡(ex)=x\ln(e^x) = x.