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Difficulty: Very hardIndices and Logarithms

Find the real value of xx that satisfies the equation log⁡2(22x+12)−x=3−log⁡2(2x−1)\log_2 (2^{2x} + 12) - x = 3 - \log_2 (2^x - 1).

  1. 1Answer
  2. B
    2
  3. C
    3
  4. D
    0

Answer

1
By bringing all terms involving xx together, the equation becomes log⁡2(22x+12)+log⁡2(2x−1)−log⁡2(2x)=3\log_2(2^{2x}+12) + \log_2(2^x-1) - \log_2(2^x) = 3. Combining these logarithmic terms yields log⁡2((22x+12)(2x−1)2x)=3\log_2\left(\frac{(2^{2x}+12)(2^x-1)}{2^x}\right) = 3. Converting to exponential form gives (22x+12)(2x−1)2x=8\frac{(2^{2x}+12)(2^x-1)}{2^x} = 8. Setting u=2xu = 2^x converts this into the cubic equation u3−u2+4u−12=0u^3 - u^2 + 4u - 12 = 0. Factoring out (u−2)(u - 2) leaves (u−2)(u2+u+6)=0(u - 2)(u^2 + u + 6) = 0. Since u2+u+6=0u^2 + u + 6 = 0 has no real solutions, u=2u = 2 is the only real root. Hence 2x=22^x = 2, which gives x=1x = 1.

Step-by-Step Solution

1
Rearrange the equation to group all logarithmic terms on one side.
\log_2 (2^{2x} + 12) + \log_2 (2^x - 1) - x = 3
Grouping logarithmic terms allows the application of logarithm combination laws.
2
Express xx as log⁡2(2x)\log_2 (2^x) and combine the logarithms using log laws.
\log_2 \left( \frac{(2^{2x} + 12)(2^x - 1)}{2^x} \right) = 3
Product and quotient laws of logarithms allow single logarithmic expression formation: log⁡2A+log⁡2B−log⁡2C=log⁡2(ABC)\log_2 A + \log_2 B - \log_2 C = \log_2\left(\frac{AB}{C}\right).
3
Convert from logarithmic to exponential form.
\frac{(2^{2x} + 12)(2^x - 1)}{2^x} = 2^3 = 8
Definition of logarithm: log⁡b(Y)=Z  ⟹  Y=bZ\log_b(Y) = Z \implies Y = b^Z.
4
Substitute u=2xu = 2^x (where u>1u > 1) to form an algebraic equation.
\frac{(u^2 + 12)(u - 1)}{u} = 8 \implies (u^2 + 12)(u - 1) = 8u
Since 22x=(2x)2=u22^{2x} = (2^x)^2 = u^2, substitution simplifies the exponential expression into a polynomial.
5
Expand and simplify into a standard cubic equation.
u^3 - u^2 + 12u - 12 = 8u \implies u^3 - u^2 + 4u - 12 = 0
Standard polynomial form enables root searching via polynomial division and factoring.
6
Factor the cubic equation.
(u - 2)(u^2 + u + 6) = 0
Testing u=2u = 2 gives 23−22+4(2)−12=8−4+8−12=02^3 - 2^2 + 4(2) - 12 = 8 - 4 + 8 - 12 = 0, confirming (u−2)(u - 2) is a factor.
7
Solve for real values of uu and xx.
u = 2 \implies 2^x = 2 \implies x = 1
The quadratic factor u2+u+6=0u^2 + u + 6 = 0 has discriminant Δ=1−24=−23<0\Delta = 1 - 24 = -23 < 0 (no real roots). Thus u=2u = 2 is the unique solution.

Key Concept

Logarithmic equations requiring exponential substitution and polynomial factorization
Estimated Time:2m 30s
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Topics

Indices and Logarithms

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