Exercise 11 Page 303

To solve the given exponential equation, we will start by rewriting the terms so that they have a common base. Note that 64 cannot be represented as a natural power of 16. Therefore, we will write both 64 and 16 as the powers of 4. Now, we have two equivalent expressions with the same base. If both sides of the equation are equal, the exponents must also be equal. 4^(6x+12) = 4^(10x) ⇔ 6x+12 = 10x Finally, we will solve the equation 6x+12 = 10x.

6x+12 = 10x

SubEqn

LHS-10x=RHS-10x

-4x+12 = 0

SubEqn

LHS-12=RHS-12

-4x = -12

DivEqn

.LHS /(-4).=.RHS /(-4).

x=-12/-4

DivNegNeg

- a/- b=a/b

x=12/4

CalcQuot

Calculate quotient

x=3

To check our answer, we will substitute 3 for x in the given equation.

64^(2x+4) = 16^(5x)

▼

x= 3

64^{2({\color{#0000FF}{3}})+4} \stackrel{?}{=} 16^{5({\color{#0000FF}{3}})}

Multiply

64^{6+4} \stackrel{?}{=} 16^{15}

Add terms

64^{10} \stackrel{?}{=} 16^{15}

Write as a power

\left( 4^3 \right)^{10} \stackrel{?}{=} \left( 4^2 \right)^{15}

'"`UNIQ--MLMath-21-QINU`"'

4^{3\cdot10} \stackrel{?}{=} 4^{2\cdot15}

Multiply

4^(30) = 4^(30) ✓

Since substituting 3 for x in the given equation produces a true statement, x=3 is the solution to our equation.