Core Connections Integrated III, 2015
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Core Connections Integrated III, 2015 View details
1. Section 9.1
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Exercise 68 Page 452

Practice makes perfect
a We want to solve an equation involving more than one logarithm. Let's use the Product Property of Logarithms to simplify the expression.
log_b mn = log_b m + log_b n We can use this rule to rewrite the left-hand side as one logarithm and solve for x!
log_2(x) + log_2(x-2) = 3

log_2(m) + log_2(n)=log_2(mn)

log_2(x * (x-2)) = 3
log_2(x^2 - 2x) = 3
We will use the Inverse Property of Logarithms to continue solving. This rule says that a logarithm and a power with the same base undo each other. We can use this property to eliminate the logarithm in our equation!
log_2(x^2 - 2x) = 3

2^(LHS)=2^(RHS)

2^(log_2(x^2 - 2x)) = 2^3

2^(log_2(m))=m

x^2 - 2x = 2^3
x^2 - 2x=8
x^2 - 2x - 8 = 0
We are left with a quadratic equation. Let's use the Quadratic Formula to find its roots. We need to identify the values of a, b, and c. x^2-2x-8=0 ⇕ 1x^2+( -2)x+( - 8)=0 We can see that a= 1, b= -2, and c= -8. Let's substitute these values into the Quadratic Formula and simplify.
x=- b±sqrt(b^2-4ac)/2a
x=- ( -2)±sqrt(( -2)^2-4( 1)( - 8))/2( 1)
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Simplify right-hand side
x=- (-2)±sqrt(4-4(1)(- 8))/2(1)
x=2±sqrt(4+32)/2
x=2±sqrt(36)/2
x = 2± 6/2
We found that x = 2 ± 62. Let's find the exact values of x.
x=2+6/2 x=8/2 x=4
x=2-6/2 x=-4/2 x=-2

We found that x can equal 4 or -2. Since we are dealing with logarithmic functions, we need to check the domain. Recall the equation before any transformations. log_2(x) + log_2(x-2) = 3 The argument of a logarithmic function must be positive. This means that arguments of both logarithms from our equation have to be positive. The solutions of these inequalities form the domain of our equation. Domain of the Equation: x > 0 and x-2 > 0 This means that x > 0 and x > 2, which simplifies into x > 2. Now let's check if the solutions are in the domain. 4 > 2 ✓ - 2 > 2 * Since x=-2 is not in our domain, we know that it is an extraneous solution. The solution of our equation is x = 4.

b We want to solve the given logarithmic equation.
log(2x)-logx^2=-2 The left-hand side of the equation involves logarithms and subtraction. This means that we can use the Quotient Property of Logarithms to simplify it. log_b(m) - log_b(n)=log_b(m/n)Let's use it to rewrite the left-hand side of the equation as one logarithm.
log(2x) - log(x^2) = - 2

log(m) - log(n)=log(m/n)

log(2x/x^2) = -2
Remember that log represents a logarithm with base 10. Let's apply the Inverse Property of Logarithms to our equation. This will eliminate the logarithm and let us continue solving.
log(2x/x^2) = -2

10^(LHS)=10^(RHS)

10^(log( 2xx^2)) = 10^(-2)

10^(log_(10)(m))=m

2x/x^2 = 10^(-2)
Let's solve the equivalent equation!
2x/x^2 =10^(-2)
2x/x^2 =1/10^2
2/x = 1/10^2
2/x = 1/100
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Solve for x
2/x* 100 = 1/100 * 100
2/x* 100=1
2* 100/x=1
200/x=1
200/x* x = x
200 = x
x=200
We have to check the domain again. The arguments of logarithms in our equation are 2x and x^2. These must be positive because a logarithmic function is defined for positive inputs only. Domain of the Equation: 2x > 0 and x^2 > 0 This means that x > 0 is the domain of our equation. Since 200 is greater than 0, x = 200 is a solution to the equation!