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Exponential Functions Simplified Revision Notes

Revision notes with simplified explanations to understand Exponential Functions quickly and effectively.

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Exponential Functions

What is an Exponential Function?

An exponential function is a mathematical function in the form:

f(x)=a×bxf(x) = a \times b^x

Where:

  • aa is the initial value (the function's value when x=0x = 0).
  • bb is the base. For exponential growth, b>1b > 1. For exponential decay, 0<b<10 < b < 1
  • xx is the exponent, usually a real number.

Key Characteristics

Domain and Range:

  • Domain: All real numbers (xRx \in \mathbb{R})
  • Range: Positive real numbers (f(x)>0f(x) > 0 if a>0a > 0)

Graph Features:

  • Exponential growth curves upwards as xx increases (b>1b>1)
  • Exponential decay curves downwards as xx increases (0<b<10 < b < 1)
  • The graph passes through the point (0,a)(0, a), since f(0)=a×b0=af(0) = a \times b^0 = a

Asymptote:

  • The xaxisx-axis (y=0y = 0) is a horizontal asymptote.

Rate of Change:

  • The rate of change increases (for growth) or decreases (for decay) exponentially.

Integration and Applications

Integration allows us to compute areas under exponential curves and model cumulative effects.

  • For exe^x
exdx=ex+C\int e^x \, dx = e^x + C
  • For a×bxa \times b^x
a×bxdx=aln(b)×bx+Cwhere b>0,b1\int a \times b^x \, dx = \frac{a}{\ln(b)} \times b^x + C \quad \text{where } b > 0, b \neq 1

These are essential for solving problems in biology, finance, and physics.


Examples

infoNote

Example 1: Growth Function


Problem:

A population of bacteria doubles every 3 hours. Initially, there are 100 bacteria. Find the population after 6 hours.


Solution:

Step 1: Start with the exponential growth formula:

P(t)=P0×btP(t) = P_0 \times b^t

Step 2: Identify the parameters:

  • P0=100P_0 = 100 (initial population).
  • Since the population doubles every 3 hours, b=213b = 2^{\frac{1}{3}}

Step 3: Write the formula:

P(t)=100×2t3P(t) = 100 \times 2^{\frac{t}{3}}

Step 4: Substitute t=6t = 6 into the formula:

P(6)=100×263=100×22=100×4P(6) = 100 \times 2^{\frac{6}{3}} = 100 \times 2^2 = 100 \times 4

Step 5: Simplify:

P(6)=400P(6) = 400

Answer: The population after 6 hours is 400.


infoNote

Example 2: Decay Function


Problem:

A radioactive substance decreases by 5% per year. Initially, there are 500 grams. Find the amount left after 10 years.


Solution:

Step 1: Start with the exponential decay formula:

A(t)=A0×btA(t) = A_0 \times b^t

Step 2: Identify the parameters:

  • A0=500A_0 = 500 (initial amount).
  • The decay rate is 55%, so b=10.05=0.95b = 1 - 0.05 = 0.95

Step 3: Write the formula:

A(t)=500×0.95tA(t) = 500 \times 0.95^t

Step 4: Substitute t=10t = 10 into the formula:

A(10)=500×0.9510A(10) = 500 \times 0.95^{10}

Step 5: Use a calculator to evaluate 0.95100.95^{10}

0.95100.59870.95^{10} \approx 0.5987

Step 6: Multiply:

A(10)=500×0.5987=299.35A(10) = 500 \times 0.5987 = 299.35

Answer: Approximately 299.35 grams remain after 10 years.


Summary

  • Exponential functions are of the form .
f(x)=a×bxf(x) = a \times b^x
  • They exhibit exponential growth (b>1b > 1) or decay (0<b<10 < b < 1).
  • The xaxisx-axis (y=0y = 0) is a horizontal asymptote.
  • Integration:
    • exdx=ex+C\int e^x dx = e^x + C
    • a×bxdx=aln(b)×bx+C\int a \times b^x dx = \frac{a}{\ln(b)} \times b^x + C
  • Applications include population growth, radioactive decay, and compound interest.
  • Key formulas:
P(t)=P0×bt(growth)P(t) = P_0 \times b^t \quad \text{(growth)} A(t)=A0×bt(decay)A(t) = A_0 \times b^t \quad \text{(decay)}
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