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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 15 Page 303

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.

1/128 = 2^(5x+3)

Write as a power

1/2^7 = 2^(5x+3)

FracToNegExponent

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

2^(-7) = 2^(5x+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. 2^(-7) = 2^(5x+3) ⇔ - 7 = 5x+3 Finally, we will solve the equation - 7 = 5x+3.

- 7 = 5x+3

LHS-3=RHS-3

- 10 = 5x

.LHS /5.=.RHS /5.

-10/5 = x

MoveNegNumToFrac

Put minus sign in front of fraction

-10/5 = x

Calculate quotient

-2 = x

Rearrange equation

x = -2

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

1/128 = 2^(5x+3)

x= -2

1/128 ? = 2^(5( -2)+3)

a(- b)=- a * b

1/128 ? = 2^(-10+3)

Add terms

1/128 ? = 2^(-7)

NegExponentToFrac

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

1/128 ? = 1/2^7

Calculate power

1/128 = 1/128 ✓

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