Exercise 2 Page 338

There are many different methods we can use to solve an exponential or logarithmic equation. Let's discuss some of them.

Solving the Equations Graphically

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

x - coordinates

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

l o g x = 1

Let's graph both sides of the equation and look for the point of intersection.

In this case, the solution is $x = 1 .$

Numerical Approach

Another way to solve these type of equations is by doing a table of values. Let's consider an exponential 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 make the $x - intervals$ smaller and evaluate between the 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 again, we could repeat the process using smaller intervals until we find the solution or until we can get an approximate answer good enough for our purpose.

Solving the Equations Algebraically

We will mention two different methods to solve these type of equations algebraically — rewriting the equation as one with a more convenient form, and using the corresponding inverse function. We will explain each of them individually.

Rewriting the Equation

One way to solve exponential or logarithmic equations is to rewrite the expressions on both of the equation's sides so that the base is more convenient. For an exponential equation, we want both sides to be exponential expressions of same base. That way we can equate the exponents.

2^{x + 1} = 4 \Leftrightarrow 2^{x + 1} = 2^{2}

This allow us to solve the equivalent equation

x + 1 = 2

instead. In the case of a logarithmic equation, recall that the definition of logarithm implies the equivalence shown below.

x = lo g_{b} y \Leftrightarrow y = b^{x}

Therefore, if we can write the logarithmic equation such that one side contains the logarithm and the other side contains the equivalent exponential form, we can identify the solution right away. Let's see an example.

lo g_{2} x = 8 \Leftrightarrow lo g_{2} x = 2^{3}

In this case we can identify the solution to be

x = 3,

without doing any more calculations.

Using the Inverse Function

Sometimes, the quantity at one side cannot be rewritten as a power with a convenient base for the given exponential or logarithmic term. In some cases, we can solve the equation by using the fact that exponential and logarithmic functions are inverse functions, and thus, they undo each other.

lo g_{2} (2^{x}) = x 2^{l o g_{2} x} = x

Consider, for example, the logarithmic equation shown below.

lo g_{2} x = 3

In this case we cannot solve the equation by rewriting it, since we cannot write

3

as a power of base

2 .

But by exponentiation we can solve it without problems.

lo g_{2} x = 3

▼

Solve for

x

$2^{LHS} = 2^{RHS}$

2^{l o g_{2} x} = 2^{3}

$2^{l o g_{2} (m)} = m$

x = 2^{3}

CalcPow

Calculate power

x = 8

Now let's consider a different equation.

2^{x} = 3^{x + 1}

In this equation we cannot rewrite the expressions to have the same base, since

2

and

3

are different prime numbers. However, we can use both sides as arguments of a logarithm and use the Properties of Logarithms to solve it.

2^{x} = 3^{x + 1}

▼

Solve for

x

$lo g (LHS) = lo g (RHS)$

lo g (2^{x}) = lo g (3^{x + 1})

$lo g (a^{m}) = m \cdot lo g (a)$

x lo g (2) = (x + 1) lo g (3)

Distr

Distribute $lo g (3)$

x lo g (2) = x lo g (3) + lo g (3)

SubEqn

$LHS - x lo g (2) = RHS - x lo g (2)$

0 = x lo g (3) - x lo g (2) + lo g (3)

FactorOut

Factor out $x$

0 = x (lo g (3) - lo g (2)) + lo g (3)

$lo g (m) - lo g (n) = lo g (\frac{m}{n})$

0 = x lo g (\frac{3}{2}) + lo g (3)

SubEqn

$LHS - lo g (3) = RHS - lo g (3)$

- lo g (3) = x lo g (\frac{3}{2})

DivEqn

$LHS / lo g (\frac{3}{2}) = RHS / lo g (\frac{3}{2})$

- \frac{lo g ( 3 )}{lo g ( \frac{3}{2} )} = x

RearrangeEqn

Rearrange equation

x = - \frac{lo g ( 3 )}{lo g ( \frac{3}{2} )}

Hints

Check the answer

Practice exercises

Asses your skill

Solving the Equations Graphically

Numerical Approach

Solving the Equations Algebraically

Rewriting the Equation

Using the Inverse Function