If 4^{x+2} \cdot 8^{x-1} = 16^{x+1}, what is the value of x? — SAT

If $4^{x+2} \cdot 8^{x-1} = 16^{x+1}$ , what is the value of $x$ ?

Answer: 3

To solve the equation

4^{x+2} \cdot 8^{x-1} = 16^{x+1}

, rewrite all bases in terms of base 2:

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

. Simplifying using the power rule yields

2^{2x+4} \cdot 2^{3x-3} = 2^{4x+4}

. Applying the product rule on the left side gives

2^{(2x+4)+(3x-3)} = 2^{5x+1}

. Equating the exponents gives

5x + 1 = 4x + 4

. Solving for

x

results in

x = 3

Express the bases 4, 8, and 16 as powers of 2.

4^{x+2} = (2^2)^{x+2} = 2^{2x+4}

8^{x-1} = (2^3)^{x-1} = 2^{3x-3}

, and

16^{x+1} = (2^4)^{x+1} = 2^{4x+4}

Writing all parts of the equation with a common base allows the exponents to be compared directly.

Combine the terms on the left side by adding their exponents.

2^{2x+4} \cdot 2^{3x-3} = 2^{(2x+4) + (3x-3)} = 2^{5x+1}

According to the product rule of exponents,

b^m \cdot b^n = b^{m+n}

when the bases are the same.

Equate the exponents from both sides of the equation.

5x + 1 = 4x + 4

If two exponential expressions with the same positive base (other than 1) are equal, their exponents must be equal.

Solve the linear equation for

x

x = 3

Subtract

4x

and

1

from both sides to isolate the variable

x

Solving exponential equations by finding a common base and applying exponent laws.

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