If 2^{x + 3} \cdot 4^{x - 1} = 16^x, what is the value of x? — GRE General Test

If $2^{x + 3} \cdot 4^{x - 1} = 16^x$ , what is the value of $x$ ?

Answer: 1

The value of

x

is 1.

Rewriting

4^{x-1}

2^{2x-2}

and

16^x

2^{4x}

transforms the left side into

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

. Setting exponents equal gives

3x + 1 = 4x

, which simplifies to

x = 1

Express all terms with a common base of 2

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

Converting

4

2^2

and

16

2^4

allows all terms to share the base 2.

Apply the power of a power rule

(a^m)^n = a^{m \cdot n}

and the product rule

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

2^{3x+1} = 2^{4x}

Multiplying exponents gives

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

and

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

. Adding exponents on the left gives

(x+3) + (2x-2) = 3x+1

Equate the exponents and solve for

x

x = 1

Since

2^A = 2^B

implies

A = B

, setting

3x + 1 = 4x

directly yields

x = 1

Solving exponential equations using common bases and exponent properties

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