To solve the given logarithmic equation, we will first recall the definition of a logarithm.

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

This tells us how we can rewrite the logarithm equivalent to

y

as an exponential equation. The argument

x

is equal to

b

raised to the power of

y .

The base of a natural logarithm is

e

, so

ln x

=

lo g_{e} x .

To use this definition, we need to rewrite the given equation.

2 ln 4 x + 5 = 8

$LHS - 5 = RHS - 5$

2 ln 4 x = 3

$b \cdot ln (a) = ln (a^{b})$

ln (4 x)^{2} = 3

Now, let's rewrite this equation in exponential form.

l n (4 x)^{2} = 3 \Leftrightarrow (4 x)^{2} = e^{3}

We will solve this equation for

x .

Let's do it!

(4 x)^{2} = e^{3}

$LHS = RHS$

4 x = \pm e^{3}

▼

Solve for

x

Write as a sum

4 x = \pm e^{1 + 2}

$a^{1 + m} = a \cdot a^{m}$

4 x = \pm e \cdot e^{2}

Commutative Property of Multiplication

4 x = \pm e^{2} \cdot e

$a \cdot b = a \cdot b$

4 x = \pm e^{2} \cdot e

$a^{2} = \pm a$

4 x = \pm e e

$LHS / 4 = RHS / 4$

x = \frac{\pm e e}{4}

The exact solutions are

x_{1} = \frac{e e}{4}

and

x_{2} = \frac{- e e}{4} .

We can also write them in decimal form using a calculator.

x_{1} x_{2} = = \frac{e e}{4} \frac{- e e}{4} \approx 1.120 \approx - 1.120

Exercise 78 Page 971

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