There are different ways to solve exponential and logarithmic equations. Just as with other type of equations, there is a graphical, numerical, and an analytical approach. We will explain an example method for each case.

Graphical Approach

To solve an exponential or logarithmic equation by graphing, we can graph both sides of the equation, each of them as an independent function. The solutions are the

x - coordinates

of the intersections points. For example, consider the equation shown below.

(\frac{1}{2})^{x} = 8

Now we graph both sides of the equation and look for the point of intersection.

The solution to this equation is $x = - 3 .$

Numerical Approach

Another way to solve these type of equations is by doing a table of values. Let's consider a different equation this time.

1 6^{x} - 2 = 0

$x$	$1 6^{x} - 2$
$- 1$	$1 6^{- 1} - 2 = - 1.9375$
$- 0.5$	$1 6^{- 0.5} - 2 = - 1.75$
$0$	$1 6^{0} - 2 = - 1$
$0.5$	$1 6^{0.5} - 2 = 2$
$1.5$	$1 6^{1.5} - 2 = 62$
$2$	$1 6^{2} - 2 = 254$

We can see that the solution happens somewhere between $0$ and $0.5,$ as there is a change of sign from one value to the other. To give a more precise answer we can do the $x - intervals$ smaller and evaluate between those values where we know our answer should be.

$x$	$1 6^{x} - 2$
$0$	$1 6^{0} - 2 = - 1$
$0.25$	$1 6^{0.25} - 2 = 0$
$0.5$	$1 6^{0.5} - 2 = 2$

As we can see, the solution to our equation is $x = 0.25 .$ If we had not found the solution once more, we can repeat the process using smaller intervals until we find the solution, or until we can give an approximate answer that is good enough for our purpose.

Analytical Approach

Another way to solve this equations is algebraically, using the fact that exponential functions and logarithmic functions are inverse functions, and therefore, they undo each other. Let's consider once more the same equation used before.

1 6^{x} - 2 = 0

To solve this equation we can isolate the exponential term and then use the corresponding logarithmic function to undue the exponential function.

1 6^{x} - 2 = 0

AddEqn

$LHS + 2 = RHS + 2$

1 6^{x} = 2

$lo g_{16} (LHS) = lo g_{16} (RHS)$

lo g_{16} 1 6^{x} = lo g_{16} 2

$lo g_{16} (16^{m}) = m$

x = lo g_{16} 2

UseCalc

Use a calculator

x = 0.25

Recall that if your calculator cannot evaluate logarithms of arbitrary base you can use the change of base formula to evaluate it in terms of common logarithms or natural logarithms.

Exercise