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Big Ideas Math Integrated I, 2016

BI

Big Ideas Math Integrated I, 2016 View details

4. Solving Exponential Equations

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Exercise 18 Page 303

Rewrite the terms so that they have a common base.

x=-10

Practice makes perfect

To solve the given exponential equation, we will start by rewriting the terms so that they have a common base.

(1/27)^(4-x) = 9^(2x-1)

Write as a power

(1/3^3)^(4-x) = (3^2)^(2x-1)

FracToNegExponent

1/a^m=a^(- m)

(3^(-3))^(4-x) = (3^2)^(2x-1)

(a^m)^n=a^(m* n)

3^(-3(4-x)) = 3^(2(2x-1))

Distribute -3

3^(-12+3x) = 3^(2(2x-1))

Distribute 2

3^(-12+3x) = 3^(4x-2)

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^(-12+3x) = 3^(4x-2) ⇔ -12+3x = 4x-2 Finally, we will solve the equation -12+3x = 4x-2.

-12+3x = 4x-2

LHS-3x=RHS-3x

-12 = x-2

LHS+2=RHS+2

-10 = x

Rearrange equation

x = -10

To check our answer, we will substitute -10 for x in the given equation.

(1/27)^(4-x) = 9^(2x-1)

x= -10

(1/27)^(4-( -10)) ? = 9^(2( -10)-1)

- (- a)=a

(1/27)^(4+10) ? = 9^(2(-10)-1)

a(- b)=- a * b

(1/27)^(4+10) ? = 9^(-20-1)

Add and subtract terms

(1/27)^(14) ? = 9^(-21)

1/a=a^(- 1)

(27^(-1))^(14) ? = 9^(-21)

(a^m)^n=a^(m* n)

27^(-1(14)) ? = 9^(-21)

Multiply

27^(-14) ? = 9^(-21)

Write as a power

(3^3)^(-14) ? = (3^2)^(-21)

(a^m)^n=a^(m* n)

3^(3(-14)) ? = 3^(2(-21))

a(- b)=- a * b

3^(-42) = 3^(-42) ✓

Since substituting -10 for x in the given equation produces a true statement, x=-10 is the solution to our equation.

Subchapter links

Communicate Your Answer p.299

Monitoring Progress p.300-302

Exercises p.303-304