Exercise 39 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. 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 it.

log_3 (x^2-12x+27) = log_3 9

Equate arguments

x^2-12x+27 = 9

SubEqn

LHS-9=RHS-9

x^2-12x+18 = 0

We will use the Quadratic Formula to solve the obatained 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^2-12x+18=0 ⇕ 1x^2+( -12)x+ 18=0 We see that a= 1, b= -12, and c= 18. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( -12)±sqrt(( -12)^2-4( 1)( 18))/2( 1)

▼

Solve for x and Simplify

NegNeg

- (- a)=a

x=12±sqrt((-12)^2-4(1)(18))/2(1)

CalcPow

Calculate power

x=12±sqrt(144-4(1)(18))/2(1)

MultByOne

a * 1=a

x=12±sqrt(144-4(18))/2

Multiply

Multiply

x=12±sqrt(144-72)/2

SubTerm

Subtract term

x=12±sqrt(72)/2

SplitIntoFactors

Split into factors

x=12±sqrt(36*2)/2

SqrtProd

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

x = 12 ± sqrt(36)*sqrt(2)/2

CalcRoot

Calculate root

x = 12 ± 6*sqrt(2)/2

FactorOut

Factor out 6

x = 6(2 ± sqrt(2))/2

ReduceFrac

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

x = 3(2±sqrt(2))

Distr

Distribute 3

x = 6 ± 3sqrt(2)

Using the Quadratic Formula, we found that the solutions of the given equation are x=6 ± 3sqrt(2). Therefore, the exact solutions are x_1=6 + 3sqrt(2) and x_2=6 - 3sqrt(2). Let's use a calculator to approximate these values. x_1 & = 6 + 3sqrt(2) ≈ 10.24 x_2 & = 6 - 3sqrt(2) ≈ 1.76 To check for extraneous solutions, we will substitute both 10.24 and 1.76 for x in the given equation one at a time.

log_3 (x-9) + log_3(x-3) = 2

Substitute

x= 10.24

log_3 ( 10.24-9) + log_3( 10.24-3) ? = 2

SubTerms

Subtract terms

log_3 1.24 + log_3 7.24 ? = 2

Calculate logarithm

0.2 + 1.8 ? = 2

AddTerms

Add terms

2 = 2 ✓

Since substituting 10.24 for x in the given equation produces a true statement, x=10.24 is a solution to our equation. Let's now check for the x=1.76.

log_3 (x-9) + log_3(x-3) = 2

Substitute

x= 1.76

log_3 ( 1.76-9) + log_3( 1.76-3) ? = 2

SubTerms

Subtract terms

log_3 (-7.24) + log_3 (-1.24) ? = 2 *

The argument of a logarithmic function cannot be negative, so both log_3 (-7.24) and log_3 (-1.24) are undefined, and 1.76 is an extraneous solution.