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# Exponents and Logarithms

Top
 Sub Topics Exponents are the numbers or variables which are raised to either constants or variables. It is also a repeated multiplication.Defined as: a$^{b}$ = a * a * a * ..... * a (b times). Functions which are distinct in the power of ‘e’ is said to be exponential function. The value of exponential function is approximately 2.718281828.Logarithm of a given number is the exponent through which a number, the base, need to be calculated to get that number. Logarithms are known as powers, and is denoted as log$_{b}(a)$ represents the power that, when we put on the base "b", gives "a". In logarithmic equations, we evaluate the unknown variable which involves log in its equation. Logarithms and exponents are also called transcendental functions because these functions transcend our ability to define them with a finite number of algebraic expressions.

## Logarithm and Exponential Rules

Given below are the important rules of exponents and logarithms.
Exponential Rules

$x^{n}.x^{m}$ = $x^{n + m}$

$x^{m-n}$ = $\frac{x^{m}}{x^{n}}$, $m$ > $n$

$(x.y)^{n}$ = $x^{n}$ . $y^{n}$

$\frac{x^{m}}{x^{n}}$ = $\frac{1}{x^{n-m}}$

($x^{n})^{m}$ = $x^{nm}$

Logarithmic Rules

ln e$^{x}$ = $x$

e$^{lnx}$ = $x$, $x$ > 0

log$_{a}$x = $\frac{log_{m}x}{log_{m}a}$

log$_{m}$1 = 0 as m$^{0}$ = 1

log$_{m}$a = p as m$^{p}$ = a

log$_{m}m^{x}$ = x as m$^{x}$ = m$^{x}$

log$_{m}$xy = log$_{m}$x + log$_{m}$y

log$_{m}\frac{x}{y}$ = log$_{m}$x - log$_{m}$y

log$_{m}x^{n}$ = n (log$_{m}$x)

## Solving Exponential Equations

There are different ways to solve the exponential equations. Some of the exponential equation problems are solved below.
Example 1: Solve 3$^{x}$ = 5

Solution: For the given problem we need to find the value of x.

Take log on both sides

log 3$^{x}$ = log 5

$x$ log 3 = log 5

$x$ = $\frac{log\ 5}{log\ 3}$

$x$ = 1.465

Example 2: Solve e$^{x}$ = 45

Solution: To find the value of x we need to use "ln".
As natural log is the inverse of exponential function.

ln e$^{x}$ = ln 45

$x$ = ln(45)

$x$ = 3.8067

Now e$^{3.807}$ = 45.0151 $\sim$ 45

Example 3: Solve 5(3)$^{x+2}$ = 25

Solution: (3)$^{x+2}$ = 5

Take log on both sides

log (3)$^{x+2}$ = log 5

($x$ + 2) log 3 = log 5

$x$ + 2 = $\frac{log\ 5}{log\ 3}$

$x$ = 1.465 - 2

$x$ = - 0.535

Example 4: Solve 3$^{x+2}$ = 3$^{5}$

Solution: For the given problem we see that the bases are same so it is easy to equate the exponents.
$x$ + 2 = 5

$x$ = 5 - 2

$x$ = 3
Therefore, the value of the variable $x$ is 3

## Derivatives of Exponential and Logarithmic Functions

Given below are some of the derivatives of exponential and logarithmic functions.

$\frac{d}{dx}$ e$^{x}$ = $e^{x}$

$\frac{d}{dx}$(log$_{e}$x) = $\frac{1}{x}$

$\frac{d}{dx}$(e$^{f(x)}$) = $f '(x)$ e$^{f(x)}$

$\frac{d}{dx}$ (ln $f(x)$) = $\frac{f'(x)}{f(x)}$

$\frac{d}{dx}$ log $x$ = $\frac{1}{x}$

e$^{ln(x)}$ = $x$ for $x$ > 0 and ln(e$^{(x)})$ = $x$ for all real $x$

## Graphing Exponential and Logarithmic Functions

Graphing exponential and logarithmic functions is pretty simple, most of the log graphs tend to have the same shape as a square-root graph. Let us explain with the help of examples:

Example 1: Graph the function $f(x)$ = e$^{2x}$

Solution:

The function e$^{2x}$ is always positive and is located in quadrants 1 and quadrants 2.
As the value of $x$ increases the value of $f(x)$ also increases. The function e$^{2x}$ is an increasing function.

Example 2: Graph the function $f(x)$ = ln (5x)

Solution:

The function $f(x)$ = ln (5x) is located in quadrants 1 and 4. Notice that the graph is not touching y axis.
This function is always positive and $x$ can never equal zero. The graph will cross the $x$ axis at 1. This is also an increasing function.

## Applications of Exponential and Logarithmic Functions

Applications of exponential and logarithmic functions can be seen in finance, physics, natural sciences etc., Some predefined rules are used to solve them.

For example: Solve 18 = 2 (3y) – 2

Move all the terms without $y$ in their exponent to one side of the equation. So, our equation now becomes: 20 = 2 (3y).
Taking log on both sides we get,

If you take log base 10, 20 = 2 (3y) becomes log10 (10 $\times$ 2) = log10 (2 (3y)).

Simplifying equation by using:

Log (ab) = b log (a) and log10 (a $\times$ b) = log10 (a) + log10 (b). We get,

=> Log10 (2) + Log10 (10) = log10 (2) + y (log10 3)

Or $y$ = $\frac{1}{ log\ 3}$