Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
2. Section 8.2
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Exercise 78 Page 393

Practice makes perfect
a When bases are not the same, we can solve an exponential equation by taking the logarithm of each side of the equation.
m=n ⇔ log m = log n Note that in order to take their logarithms both m and n must be positive numbers. In our case, we need to first isolate the exponential expression. Then we will solve our equation using the Properties of Logarithms.
5.2(3.75)^x = 100
(3.75)^x = 100/5.2

log(LHS)=log(RHS)

log (3.75)^x = log 100/5.2

log(a^m)= m*log(a)

x log 3.75 = log 100/5.2
x=log 1005.2/log 3.75
x≈ 2.2368

Round to 2 decimals

x≈ 2.24
b We want to solve the given polynomial equation. To do so, we will apply a factoring technique called the difference of two squares.
a^2-b^2 ⇔ (a+b)(a-b) To do so, we will start by simplifying the equation to obtain the difference of two terms. Then, we will write both terms as perfect squares.
4+3x^4=81
3x^4 = 77
x^4 = 77/3
x^4 - 77/3 = 0
x^(2*2) - 77/3 = 0
( x^2 ) ^2-77/3=0

a = ( sqrt(a) )^2

( x^2 ) ^2-(sqrt(77/3))^2=0
( x^2+sqrt(77/3) ) ( x^2-sqrt(77/3) ) =0
We have written the left-hand side of the equation as the product of two factors. To solve the equation, we will apply the Zero Product Property.
( x^2+sqrt(77/3) ) ( x^2-sqrt(77/3) ) =0
lcx^2+sqrt(77/3)=0 & (I) x^2-sqrt(77/3)=0 & (II)
â–¼
(I), (II): Solve for x
lx^2=-sqrt(77/3) x^2-sqrt(77/3)=0
lx^2=-sqrt(77/3) x^2=sqrt(77/3)

(I), (II): sqrt(LHS)=sqrt(RHS)

lx=±sqrt(-sqrt(77/3)) x=±sqrt(sqrt(77/3))
lx=± isqrt(sqrt(77/3)) x=±sqrt(sqrt(77/3))

(I), (II): sqrt(a)=a^(12)

lx=± i((77/3)^(12))^(12) x=±((77/3)^(12))^(12)

(I), (II): (a^m)^n=a^(m* n)

lx=± i(77/3)^(12* 12) x=±(77/3)^(12* 12)

(I), (II):Multiply fractions

lx=± i(77/3)^(14) x=±(77/3)^(14)

(I), (II): a^(1n)=sqrt(a)

lx=± isqrt(77/3) x=±sqrt(77/3)
lx=±sqrt(77/3)i x=±sqrt(77/3)

(I), (II): Use a calculator

lx≈± 2.25i x≈± 2.25
We found four approximate solutions for the given equation. 2.25, - 2.25, 2.25i, - 2.25i