Exponential Function Formulas

The exponential function, typically denoted as e^x or exp⁡(x), is a mathematical function where 'e' is the base of the natural logarithm, approximately equal to 2.71828. The formula for the exponential function is:

f(x) = a^x

This function is characterized by the fact that the rate of growth is proportional to the value of the function. That is, the derivative of e^x concerning x is e^x itself. This unique property makes the exponential function very important in fields such as mathematics, physics, engineering, and finance.

An exponential function is a mathematical function of the shape f (x) = a^x, where 'x' is variable and 'a' is consistent this is the function's base and needs to be more than 0.

'e' is a transcendental number, which is about the same as 2.71828, and is the most customarily used exponential function basis.

For Examples,

• f(x) = 2^x
• f(x) = (1/2)^x
• f(x) = 3e^2x
• f(x) = 4 (3)^-0.5x

Below are different formulas for exponential functions.

Exponential Growth Formula

At first, the quantity grows very slowly, then rapidly in Exponential Growth. Over time, the rate of change accelerates. With time, the rate of growth accelerates. The quick expansion is described as an "exponential increase." 

The exponential growth formula is as follows:

y = a (1 + r) ^x

Exponential Decay Formula

In Exponential Decay, the quantity decreases rapidly at first, then gradually. The rate of change slows over time. The rate of change slows with time. The rapid rise was supposed to create an "exponential decline." 

The formula for exponential growth is as follows:

y = a (1 - r) ^x

Exponential Series

The following power series can be used to define the real exponential function.

e^x = ∑^∞_n=0 xⁿ/n! = (1/1) + (x/1) + (x²/2) + (x³/6) + ...

Some other exponential functions' expansions are illustrated below,

e = ∑^∞_n=0 xⁿ/n! = (1/1) + (1/1) + (1/2) + (1/6) + ...
e^-1 = ∑^∞_n=0 xⁿ/n! = (1/1) - (1/1) + (1/2) - (1/6) + ...

Exponential Function Rules

The following are some important exponential rules:

For any real numbers x and y, if a > 0 and b > 0, the following is true:

• a^x a^y = a ^{x + y}
• a^x/a^y = a^x-y
• (a^x)^y = a ^xy
• a^xb^x = (ab)^x
• (a/b)^x = a^x/b^x
• a⁰ = 1
• a^-x = 1/ a^x

Exponential Function Derivative Formula

The differentiation formulas that are used to obtain the exponential function's derivative.

d/dx (e^x) = e^x
d/dx (a^x) = a^x · ln a

Exponential Function Integration Formula

The integral of an exponential function is calculated using integration formulae.

∫ e^x dx = e^x + C
∫ a^x dx = a^x / (ln a) + C

Applications of Exponential Functions

Some applications of exponential function formulas are:

• Population Growth: Modeling populations growing over time.
• Radioactive Decay: Describing the decay of radioactive substances.
• Compound Interest: Calculating interest in finance.
• Physics and Engineering: Modeling processes like cooling and heating.

Related Reads:

• Derivative of Exponential Function
• Exponential Graph

Solved Problems on Exponential Function Formula

Problem 1: In 2010, a town had 100,000 inhabitants. How many citizens will there be in ten years if the population grows at an annual rate of 8%? 

Solution:

Initial population is, a = 100,000.
Rate of growth is, r = 8% = 0.08.
Time is, x = 10 years.

Using exponential growth formula,

f(x) = a (1 + r)^x
f(x) = 100000(1 + 0.08)¹⁰
f(x) ≈ 215,892

Problem 2: Carbon-14 has a half-life of 5,730 years. What is the quantity of carbon left after 2000 years if there were 1000 kilos of carbon at the start?

Solution:

Using the given data, we can say that carbon-14 is decaying and hence we use the formula of exponential decay.

P = P₀ e^{-k t} ... (1),

Here, P₀= initial amount of carbon = 1000 grams.

It is given that the half-life of carbon-14 is 5,730 years. It means
P = P₀/ 2 = 1000 / 2 = 500 grams.

Substitute all these values in (1),
500 = 1000 e^{-k (5730)}

Dividing both sides by 1000,
0.5 = e^{-k (5730)}

Taking "ln" on both sides,
ln 0.5 = -5730k

Dividing both sides by -5730,
k = ln 0.5 / (-5730) ≈ 0.00012097

We have to find the amount of carbon that is left after 2000 years. Substitute t = 2000 in (1),

P = 1000 e ^{-(0.00012097) (2000)} ≈ 785 grams.

Problem 3: Simplify the following exponential expression: 3^x - 3^x+2.

Solution:

Given exponential equation: 3^x - 3^x+2

By using the property: a^x a^y = a^x+y

Hence, 3^x+2 can be written as 3^x.3²

Thus the given equation is written as: 3^x - 3^x+2 = 3^x - 3^x·9

Now, factor out the term 3^x

3^x - 3^x+2 = 3^x - 3^x·9 = 3^x(1 - 9)
3^x - 3^x+2 = 3^x(-8)
3^x - 3^x+2 = -8(3^x)

Therefore, the simplification of the given exponential equation 3^x-3^x+1 is -8(3^x).

Problem 4: Solve the exponential equation: (¼)^x = 64.

Solution:

Given exponential equation is: (¼)^x = 64

Using the exponential rule (a/b)^x = a^x/b^x , we get;

1^x/4^x = 43
1/4^x = 43 [since 1^x = 1]
(1)(4^-x) = 4³
4^-x = 4³

Here, bases are equal.

So, x = -3

Problem 5: Simplify the exponential function 2^x – 2^x+1.

Solution:

Given exponential function: 2^x – 2^x+1

By using the property: a^x a^y = a ^{x + y}

Hence, 2^x+1 can be written as 2^x. 2

Thus, the given function is written as: 2^x - 2^x+1 = 2^x - 2^x. 2

Now, factor out the term 2^x

2^x - 2^x+1 = 2^x - 2^x. 2 = 2^x(1-2)
2^x - 2^x+1 = 2^x(-1)
2^x - 2^x+1 = – 2^x

Therefore, the simplification of the given exponential function 2^x – 2^x+1 is – 2^x.

Practice Problems on Exponential Function Formula

Question 1: Simplify the exponential function: y = (3⁴)² + 2.

Question 2: Find the value of x given that: (x /2)² = 4.

Question 3: Integrate the exponential function e^3x + 2.

Question 4: Differentiate the exponential function 2^x + e^4x.

Question 5: Solve: 5^x + 5^x+1 = 150.