Exercise 36 Page 339

We want to solve an equation involving more than one logarithm. To do so, we will use the Product Property of Logarithms. log_b mn = log_b m + log_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. Next, we will use the fact that if two equivalent logarithmic expressions have the same base, then the arguments must be equal. log_b x=log_b y ⇔ x= y Let's apply the above property to our equation and then simplify the equation.

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

Equate arguments

x^2-2x = e^5

SubEqn

LHS-e^5=RHS-e^5

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

We will use the Quadratic Formula to solve the given quadratic equation. ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 a We first need to identify the values of a, b, and c. x^2-2x-e^5=0 ⇕ 1x^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=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( -2)±sqrt(( -2)^2-4( 1)( - e^5))/2( 1)

▼

Solve for x and Simplify

NegNeg

- (- a)=a

x=2±sqrt((-2)^2-4(1)(- e^5))/2(1)

CalcPow

Calculate power

x=2±sqrt(4-4(1)(- e^5))/2(1)

MultByOne

a * 1=a

x=2±sqrt(4-4(- e^5))/2

MultNegNegOnePar

- a(- b)=a* b

x=2±sqrt(4+4e^5)/2

FactorOut

Factor out 4

x=2±sqrt(4(1+e^5))/2

SplitIntoFactors

Split into factors

x=2±sqrt(4*(1+e^5))/2

SqrtProd

sqrt(a* b)=sqrt(a)*sqrt(b)

x = 2± sqrt(4) * sqrt(1+e^5)/2

CalcRoot

Calculate root

x = 2± 2 * sqrt(1+e^5)/2

FactorOut

Factor out 2

x = 2(1± sqrt(1+e^5))/2

ReduceFrac

a/b=.a /2./.b /2.

x = 1± sqrt(1+e^5)

Using the Quadratic Formula, we found that the solutions of the given equation are x=1± sqrt(1+e^5). Therefore, the exact solutions are x_1=1+sqrt(1+e^5) and x_2=1-sqrt(1+e^5). Let's use a calculator to approximate these values. x_1 & = 1+sqrt(1+e^5) ≈ 13.223 x_2 & =1-sqrt(1+e^5) ≈ - 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 *

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.