Question

Difficulty: MediumExponential Functions and Equations

If 22x+222x=122^{2x+2} - 2^{2x} = 12 for some real number xx, what is the value of 24x2^{4x}?

Answer: 16

Answer

16
By applying the product rule of exponents, the expression 22x+22^{2x+2} can be rewritten as 22x222^{2x} \cdot 2^2, or 422x4 \cdot 2^{2x}. Substituting this into the given equation yields 422x22x=124 \cdot 2^{2x} - 2^{2x} = 12. Factoring out 22x2^{2x} gives 22x(41)=122^{2x}(4 - 1) = 12, which simplifies to 322x=123 \cdot 2^{2x} = 12. Dividing both sides of the equation by 3 results in 22x=42^{2x} = 4. Since 24x2^{4x} can be written as (22x)2(2^{2x})^2, substituting 4 for 22x2^{2x} gives 42=164^2 = 16. Alternatively, solving 22x=42^{2x} = 4 gives 2x=22x = 2, which means x=1x = 1. Substituting x=1x = 1 into 24x2^{4x} yields 24(1)=24=162^{4(1)} = 2^4 = 16.

Step-by-Step Solution

1
Rewrite 22x+22^{2x+2} using exponent properties.
22x+2=22x22=422x2^{2x+2} = 2^{2x} \cdot 2^2 = 4 \cdot 2^{2x}
To express both exponential terms with the same base power, allowing them to be factored or combined.
2
Substitute this back into the equation and factor out the common term 22x2^{2x}.
22x(41)=12    322x=122^{2x}(4 - 1) = 12 \implies 3 \cdot 2^{2x} = 12
To isolate the exponential expression.
3
Solve for 22x2^{2x} by dividing both sides of the equation by 3.
22x=42^{2x} = 4
To find the value of the exponential term.
4
Express 24x2^{4x} in terms of 22x2^{2x} and evaluate.
24x=(22x)2=42=162^{4x} = (2^{2x})^2 = 4^2 = 16
To find the final requested value using the exponent rule (am)n=amn(a^m)^n = a^{mn}.

Key Concept

Factoring exponential equations and applying power of a power exponent rules
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