Question

Difficulty: MediumLogarithmic and Exponential Expressions and Equations

If a>1.5a > 1.5 is a real number that satisfies the equation 4a3(2a+1)+8=04^a - 3(2^{a+1}) + 8 = 0, what is the value of aa?

Answer: 2

Answer

The correct answer is 2.
Substituting u=2au = 2^a turns the equation into u26u+8=0u^2 - 6u + 8 = 0. Factoring gives (u2)(u4)=0(u-2)(u-4)=0, meaning u=2u=2 or u=4u=4. Reversing the substitution gives 2a=2    a=12^a = 2 \implies a=1 and 2a=4    a=22^a = 4 \implies a=2. Since a>1.5a > 1.5, the correct value is 2.

Step-by-Step Solution

1
Express the equation in terms of base 2.
(2a)26(2a)+8=0(2^a)^2 - 6(2^a) + 8 = 0
Since 4 is 222^2 and 2a+1=22a2^{a+1} = 2 \cdot 2^a, expressing all terms in base 2 allows for algebraic substitution.
2
Substitute u=2au = 2^a to form a quadratic equation.
u26u+8=0u^2 - 6u + 8 = 0
Substitution simplifies the exponential equation into a standard quadratic form.
3
Factor the quadratic equation.
(u2)(u4)=0(u-2)(u-4) = 0
Factoring allows us to find the roots of the quadratic equation.
4
Solve for the variable aa.
a=1a = 1 or a=2a = 2
Solving 2a=22^a = 2 yields a=1a = 1, and solving 2a=42^a = 4 yields a=2a = 2.
5
Apply the given constraint on aa.
a=2a = 2
The problem states that a>1.5a > 1.5, so a=1a = 1 is discarded and a=2a = 2 is the correct value.

Key Concept

Solving exponential equations using quadratic substitution

Alternative Method

Instead of using substitution, test values for aa. Since a>1.5a > 1.5, testing small integer values starting with a=2a=2 shows 423(23)+8=1624+8=04^2 - 3(2^3) + 8 = 16 - 24 + 8 = 0, validating that a=2a=2 is the solution.
Estimated Time:1m 30s
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