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)

WritePow

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. \

begin{gathered} 2^{\text{-}7} = 2^{5x+3} \quad \Leftrightarrow \quad \text{-} 7 = 5x+3 \end{gathered} Finally, we will solve the equation - 7 = 5x+3.

- 7 = 5x+3

SubEqn

LHS-3=RHS-3

- 10 = 5x

DivEqn

.LHS /5.=.RHS /5.

-10/5 = x

MoveNegNumToFrac

Put minus sign in front of fraction

-10/5 = x

CalcQuot

Calculate quotient

-2 = x

RearrangeEqn

Rearrange equation

x = -2

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

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

Substitute

x= -2

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

MultPosNeg

a(- b)=- a * b

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

AddTerms

Add terms

1/128 ? = 2^(-7)

NegExponentToFrac

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

1/128 ? = 1/2^7

CalcPow

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

Exercise 15 Page 303

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