If \frac{16^{x+1}}{2^{x-3}} = 64, what is the value of x? — SAT

If $\frac{16^{x+1}}{2^{x-3}} = 64$ , what is the value of $x$ ?

-\frac{1}{3}

The correct answer is

-\frac{1}{3}

. By expressing all terms with base 2, the equation

\frac{16^{x+1}}{2^{x-3}} = 64

becomes

\frac{2^{4x+4}}{2^{x-3}} = 2^6

. Applying exponent division rules yields

2^{(4x+4)-(x-3)} = 2^6

, which simplifies to

2^{3x+7} = 2^6

. Setting the exponents equal gives

3x + 7 = 6

, which simplifies to

x = -\frac{1}{3}

Rewrite all terms in the equation using a common base of 2.

The equation becomes

\frac{(2^4)^{x+1}}{2^{x-3}} = 2^6

Expressing bases 16 and 64 as powers of 2 allows us to apply exponent laws to simplify the equation.

Apply the power of a power rule to the numerator, then apply the quotient rule for exponents.

The left side simplifies to

2^{4x+4 - (x-3)} = 2^{3x+7}

The expression

(2^4)^{x+1}

becomes

2^{4x+4}

. Using the quotient rule

\frac{a^m}{a^n} = a^{m-n}

, we subtract the denominator's exponent

(x-3)

from the numerator's exponent

(4x+4)

, distributing the negative sign to get

3x+7

Set the exponents equal to each other and solve the resulting linear equation.

3x + 7 = 6 \implies 3x = -1 \implies x = -\frac{1}{3}

Since the bases on both sides of the equation are equal, their exponents must be equal.

Solving exponential equations by rewriting with a common base

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