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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 7 Page 301

Rewrite the terms so that they have a common base.

x = -2

Practice makes perfect

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

(1/3)^(x-1) = 27

1/a=a^(- 1)

(3^(-1))^(x-1) = 27

Write as a power

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

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

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

Distribute -1

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

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^(- x+1) = 3^3 ⇔ - x+1 = 3 Finally, we will solve the equation - x+1 = 3.

- x+1 = 3

LHS-1=RHS-1

- x = 2

.LHS /(-1).=.RHS /(-1).

x=2/-1

MoveNegDenomToFrac

Put minus sign in front of fraction

x = -2/1

a/1=a

x = -2

To check our answer, we will substitute -2 for x in the given equation.

(1/3)^(x-1) = 27

x= -2

(1/3)^(-2-1) ? = 27

Subtract term

(1/3)^(-3) ? = 27

1/a=a^(- 1)

(3^(-1))^(-3) ? = 27

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

3^(-1(-3)) ? = 27

MultNegNegOnePar

- a(- b)=a* b

3^3 ? = 27

Calculate power

27 = 27 ✓

Since substituting -2 for x in the given equation produces a true statement, x=-2 is the solution to our equation.

Subchapter links

Communicate Your Answer p.299

Monitoring Progress p.300-302

Exercises p.303-304