We want to solve an equation involving more than one logarithm. To do so, we will use the Product Property of Logarithms.

lo g_{b} m n = lo g_{b} m + lo g_{b} n

First, we will isolate the logarithm that contains the variable. Then, we can use the above property to isolate the variable from the logarithm. Note that the property holds also for the natural logarithm.

ln x + ln (x - 2) = 5

$ln (a) + ln (b) = ln (a b)$

ln (x (x - 2)) = 5

Distr

Distribute $x$

ln (x^{2} - 2 x) = 5

$ln (e^{a}) = a$

ln (x^{2} - 2 x) = ln e^{5}

Next, we will use the fact that if two equivalent logarithmic expressions have the same base, then the arguments must be equal.

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

Let's apply the above property to our equation and then simplify the equation.

ln (x^{2} - 2 x) = ln e^{5}

Equate arguments

x^{2} - 2 x = e^{5}

SubEqn

$LHS - e^{5} = RHS - e^{5}$

x^{2} - 2 x - e^{5} = 0

We will use the Quadratic Formula to solve the given quadratic equation.

a x^{2} + b x + c = 0 ⇕ x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

We first need to identify the values of

a,

b,

and

c .

x^{2} - 2 x - e^{5} = 0 ⇕ 1 x^{2} + (- 2) x + (- e^{5}) = 0

We see that

a = 1,

b = - 2,

and

c = - e^{5} .

Let's substitute these values into the Quadratic Formula.

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValues

Substitute values

x = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( 1 ) ( - e ^{5} )}{2 ( 1 )}

▼

Solve for

x

and Simplify

NegNeg

$- (- a) = a$

x = \frac{2 \pm ( - 2 ) ^{2} - 4 ( 1 ) ( - e ^{5} )}{2 ( 1 )}

CalcPow

Calculate power

x = \frac{2 \pm 4 - 4 ( 1 ) ( - e ^{5} )}{2 ( 1 )}

MultByOne

$a \cdot 1 = a$

x = \frac{2 \pm 4 - 4 ( - e ^{5} )}{2}

MultNegNegOnePar

$- a (- b) = a \cdot b$

x = \frac{2 \pm 4 + 4 e ^{5}}{2}

FactorOut

Factor out $4$

x = \frac{2 \pm 4 ( 1 + e ^{5} )}{2}

SplitIntoFactors

Split into factors

x = \frac{2 \pm 4 \cdot ( 1 + e ^{5} )}{2}

SqrtProd

$a \cdot b = a \cdot b$

x = \frac{2 \pm 4 \cdot 1 + e ^{5}}{2}

CalcRoot

Calculate root

x = \frac{2 \pm 2 \cdot 1 + e ^{5}}{2}

FactorOut

Factor out $2$

x = \frac{2 ( 1 \pm 1 + e ^{5} )}{2}

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

x = 1 \pm 1 + e^{5}

Using the Quadratic Formula, we found that the solutions of the given equation are

x = 1 \pm 1 + e^{5} .

Therefore, the exact solutions are

x_{1} = 1 + 1 + e^{5}

and

x_{2} = 1 - 1 + e^{5} .

Let's use a calculator to approximate these values.

x_{1} x_{2} = 1 + 1 + e^{5} \approx 13.223 = 1 - 1 + e^{5} \approx - 11.223

To check for extraneous solutions, we will substitute both

13.223

and

- 11.223

for

x

in the given equation one at a time.

ln x + ln (x - 2) = 5

Substitute

$x = 13.223$

ln 13.223 + ln (13.223 - 2) = ? 5

SubTerm

Subtract term

ln 13.223 + ln 11.223 = ? 5

Calculate logarithm

2.582 + 2.418 = ? 5

AddTerms

Add terms

5 = 5 ✓

Since substituting

13.223

for

x

in the given equation produces a true statement,

x = 13.223

is a solution to our equation. Let's now check

x = - 11.223 .

ln x + ln (x - 2) = 5

Substitute

$x = - 11.223$

ln (- 11.223) + ln (- 11.223 - 2) = ? 5

SubTerm

Subtract term

ln (- 11.223) + ln (- 13.223) = ? 5 \times

The argument of a logarithmic function cannot be negative, so both

ln (- 11.223)

and

ln (- 13.223)

are undefined, and

- 11.223

is an extraneous solution.

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