Exponential Functions Solver

Evaluate Exponential Functions

Select the form of the exponential function and enter the parameters to calculate the value of $y$. The solver will show each step of the calculation.

Function Form:

Parameters for $y = ab^x$

$a$ (Initial Amount):

$b$ (Growth/Decay Factor):

$x$ (Exponent/Time):

Enter parameters and click "Calculate y".

Interactive Graphs

Visualize the exponential function. Parameters are linked to the solver. Calculate above or adjust here for quick visualization.

$a$ (Initial Amount):

$b$ (Growth/Decay Factor for $ab^x$):

$x_{max}$ (Maximum x-value for graph):

Understanding Exponential Functions

Exponential functions are fundamental in mathematics and describe processes where the rate of change is proportional to the current amount. This means that the larger a quantity is, the faster it grows (or the slower it decays, if the rate is negative). They are widely used to model phenomena such as:

Population growth of species.
Radioactive decay of unstable isotopes.
Compound interest earned on investments.
Spread of diseases in epidemiology.
Cooling or heating of objects.

Common Forms of Exponential Functions

Two primary forms are commonly encountered:

Form 1: $y = ab^x$
This form is often used when the growth or decay occurs in discrete intervals or is described by a factor per period.
- $y$: The final amount after $x$ periods.
- $a$: The initial amount (the value of $y$ when $x=0$). This must be non-zero for a non-trivial exponential function. If $a=0$, then $y=0$ always.
- $b$: The growth/decay factor per unit of $x$. This factor must be positive ($b > 0$) and not equal to 1 ($b \neq 1$).
  - If $b > 1$: The function represents exponential growth (e.g., doubling, $b=2$).
  - If $0 < b < 1$: The function represents exponential decay (e.g., halving, $b=0.5$).
- $x$: The exponent, often representing time, number of periods, or another independent variable.
Form 2: $y = ae^{kx}$ (The Natural Exponential Function)
This form is particularly useful for modeling processes involving continuous growth or decay.
- $y$: The final amount after time $x$.
- $a$: The initial amount (the value of $y$ when $x=0$). Similar to the first form, $a \neq 0$ for a non-trivial function.
- $e$: Euler's number, an important mathematical constant approximately equal to $2.718281828...$. It is the base of the natural logarithm.
- $k$: The continuous growth or decay rate.
  - If $k > 0$: The function represents continuous exponential growth.
  - If $k < 0$: The function represents continuous exponential decay.
  - If $k = 0$: The function becomes $y = a$, a constant function.
- $x$: The exponent, typically representing time.

Relationship between the forms: The two forms are interchangeable. If $y = ab^x$ and $y = ae^{kx}$, then $b^x = e^{kx}$. Taking the natural logarithm of both sides of $b = e^k$ gives $\ln(b) = \ln(e^k)$, which simplifies to $k = \ln(b)$. Conversely, $b = e^k$. This means you can convert between a base-$b$ factor and a continuous rate $k$.

Examples of Growth and Decay

Exponential Growth

Example 1: Population Growth ($y = ab^x$)

Imagine a bacterial culture starts with 100 cells ($a=100$). If the number of cells doubles every hour (meaning the growth factor $b=2$), the population after $x$ hours can be modeled by $y = 100 \cdot 2^x$.

After 3 hours, the population would be: $y = 100 \cdot 2^3 = 100 \cdot 8 = 800$ cells.

Example 2: Compound Interest (Continuous Compounding, $y = ae^{kx}$)

Suppose you invest $1000 ($a=1000$) in an account that earns an annual interest rate of 5% compounded continuously. The continuous growth rate $k$ is $0.05$. The amount of money $y$ after $x$ years is given by $y = 1000e^{0.05x}$.

After 10 years, the amount would be: $y = 1000e^{0.05 \cdot 10} = 1000e^{0.5} \\approx 1000 \cdot 1.64872 \\approx \$1648.72$.

Exponential Decay

Example 1: Radioactive Decay ($y = ab^x$)

Consider a radioactive substance with a half-life of 10 years. This means that every 10 years, half of the substance decays. If you start with 200 grams ($a=200$), the decay factor for one half-life period is $b=0.5$. If $x$ represents the number of 10-year periods, the amount remaining is $y = 200 \cdot (0.5)^x$.

After 3 such periods (30 years): $y = 200 \cdot (0.5)^3 = 200 \cdot 0.125 = 25$ grams.

If we want to model this with $t$ in years (not periods), we need to adjust $b$. Since $b=(0.5)$ for $x=1$ (10 years), for $t$ years, $b_{year}^{10} = 0.5$, so $b_{year} = (0.5)^{1/10} \\approx 0.93303$. The formula becomes $y = 200 \cdot (0.93303)^t$.

Example 2: Drug Clearance ($y = ae^{kx}$)

A certain drug is eliminated from the body at a continuous rate. If the initial dose is 50mg ($a=50$) and the continuous decay rate is $k = -0.2$ per hour (this means the amount is continuously decreasing at a rate equivalent to 20% per hour if it were simple decay, but it's continuous), the amount of drug $y$ remaining in the body after $x$ hours is $y = 50e^{-0.2x}$.

After 5 hours: $y = 50e^{-0.2 \cdot 5} = 50e^{-1} \\approx 50 \cdot 0.36788 \\approx 18.39$ mg.