Exercise 9 Page 303

Rewrite the terms so that they have a common base.

x=-5

Practice makes perfect

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

7^(x-5) = 49^x

WritePow

Write as a power

7^(x-5) = ( 7^2 )^x

PowPow

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

7^(x-5) = 7^(2x)

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

x-5=2x

SubEqn

LHS-2x=RHS-2x

- x-5 = 0

AddEqn

LHS+5=RHS+5

- x = 5

DivEqn

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

x=5/-1

MoveNegDenomToFrac

Put minus sign in front of fraction

x=-5/1

DivByOne

a/1=a

x = -5

To check our answer, we will substitute -5 for x in the given equation.

7^(x-5) = 49^x

Substitute

x= -5

7^(-5-5) ? = 49^(-5)

SubTerm

Subtract term

7^(-10) ? = 49^(-5)

WritePow

Write as a power

7^(-10) ? = (7^2)^(-5)

PowPow

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

7^(-10) ? = 7^(2*(-5))

MultPosNeg

a(- b)=- a * b

7^(-10) = 7^(-10) ✓

Since substituting -5 for x in the given equation produces a true statement, x=-5 is the solution to our equation.

Subchapter links

Communicate Your Answer p.299

Monitoring Progress p.300-302

Exercises p.303-304

Exercise 9 Page 303

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