Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
1. Section 11.1
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Exercise 13 Page 603

Practice makes perfect
a To solve the given exponential equation, we will start by rewriting the terms so that they have the same base.
25^2=125^(x+1)
( 5^2 )^2=( 5^3 )^(x+1)
5^4=5^(3(x+1))
Now, we have two equivalent expressions with the same base. If both sides of the equation are equal, the exponents must also be equal. 5^4=5^(3(x+1)) ⇔ 4=3(x+1) Finally, we will solve the equation 4=3(x+1).
4=3(x+1)
â–Ľ
Solve for x
4=3x+3
1=3x
1/3=x
x=1/3
b This equation would be much easier to solve if it had no fractions. We can start solving by changing this equation to a simpler equivalent equation by eliminating fractions. To do this, we will multiply both sides of the equation by 15, which is the least common denominator.
x/5+x-1/3=2
15x/5+15(x-1)/3=30
3x+5(x-1)=30
To solve an equation we should first gather all of the variable terms on one side of the equation and all of the constant terms on the other side using the Properties of Equality.
3x+5(x-1)=30
â–Ľ
Solve for x
3x+5x-5=30
3x+5x=35
8x=35
x=35/8
c To solve the given quadratic equation, we will start by paying close attention to its terms. Note that the first and last terms are perfect squares, and that the middle term is twice the product of their roots. Thus, it is a perfect square trinomial and we can factor it following the corresponding formula.
a^2+2ab+b^2=(a+b)^2 Let's solve the equation by factoring the perfect square trinomial on the left-hand side.
x^2-4x+4=25
â–Ľ
a^2-2ab+b^2=(a-b)^2
x^2-4x+2^2=25
x^2-2(x)(2)+2^2=25
(x-2)^2=25
x-2=± 5
x=2± 5
We will now find the first and second solutions by using the positive and negative signs.
x=2± 5
x_1=2+5 x_2=2- 5
x_1=7 x=- 3
d To solve the given exponential equation, we will start by rewriting the terms so that they have the same base. Recall that 1a=a^(- 1).
9^x=( 1/3 )^(x+3)
( 3^2 )^x=( 1/3 )^(x+3)

1/a=a^(- 1)

( 3^2 )^x=( 3^(- 1) )^(x+3)
3^(2x)=3^(- x -3)
Now, we have two equivalent expressions with the same base. If both sides of the equation are equal, the exponents must also be equal. 3^(2x)=3^(- x -3) ⇔ 2x=- x-3 Finally, we will solve the equation 2x=- x-3.
2x=- x-3
â–Ľ
Solve for x
3x=- 3
x=- 1