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Difficulty: Very hardLogarithmic and Exponential Expressions and Equations

If xx is a positive real number such that log⁡2(x)⋅log⁡3(x)−log⁡2(x)−log⁡3(x)=1\log_2(x) \cdot \log_3(x) - \log_2(x) - \log_3(x) = 1, what is the product of all possible real values of xx?

  1. A
    16\frac{1}{6}
  2. B
    55
  3. 66Answer
  4. D
    15\frac{1}{5}
  5. E
    1212

Answer

6
The correct answer is 6. By converting the equation using the change-of-base formula into natural logarithms, we obtain the quadratic equation (ln⁡x)2−(ln⁡2+ln⁡3)ln⁡x−ln⁡2⋅ln⁡3=0(\ln x)^2 - (\ln 2 + \ln 3)\ln x - \ln 2 \cdot \ln 3 = 0. By setting w=ln⁡xw = \ln x, we have a quadratic in terms of ww. The product of the two solutions x1x_1 and x2x_2 is x1x2=ew1ew2=ew1+w2x_1 x_2 = e^{w_1} e^{w_2} = e^{w_1 + w_2}. Using Vieta's formulas, the sum of the roots w1+w2=ln⁡2+ln⁡3=ln⁡6w_1 + w_2 = \ln 2 + \ln 3 = \ln 6. Therefore, the product of the solutions is eln⁡6=6e^{\ln 6} = 6.

Step-by-Step Solution

1
Rewrite the logarithmic equation using a common base.
Using the change-of-base formula log⁡b(a)=ln⁡aln⁡b\log_b(a) = \frac{\ln a}{\ln b}, the equation becomes: (ln⁡xln⁡2)(ln⁡xln⁡3)−ln⁡xln⁡2−ln⁡xln⁡3=1\left(\frac{\ln x}{\ln 2}\right)\left(\frac{\ln x}{\ln 3}\right) - \frac{\ln x}{\ln 2} - \frac{\ln x}{\ln 3} = 1
This allows all logarithmic terms to be expressed in terms of the natural logarithm, making it easier to solve.
2
Eliminate denominators by multiplying the equation by ln⁡2⋅ln⁡3\ln 2 \cdot \ln 3.
(ln⁡x)2−(ln⁡3)ln⁡x−(ln⁡2)ln⁡x=ln⁡2⋅ln⁡3(\ln x)^2 - (\ln 3)\ln x - (\ln 2)\ln x = \ln 2 \cdot \ln 3
Multiplying through by the common denominator simplifies the rational equation into a standard polynomial form.
3
Rearrange the equation into a standard quadratic form.
(ln⁡x)2−(ln⁡2+ln⁡3)ln⁡x−ln⁡2⋅ln⁡3=0(\ln x)^2 - (\ln 2 + \ln 3)\ln x - \ln 2 \cdot \ln 3 = 0
Grouping the coefficients of ln⁡x\ln x reveals a quadratic equation of the form Aw2+Bw+C=0Aw^2 + Bw + C = 0, where w=ln⁡xw = \ln x.
4
Define the relationship between the roots of the quadratic equation and the values of xx.
Let the two real roots of the quadratic equation be w1w_1 and w2w_2. These correspond to the solutions for xx, namely x1=ew1x_1 = e^{w_1} and x2=ew2x_2 = e^{w_2}. The product of the solutions is x1⋅x2=ew1⋅ew2=ew1+w2x_1 \cdot x_2 = e^{w_1} \cdot e^{w_2} = e^{w_1 + w_2}.
To find the product of the solutions for xx, we must compute the exponential of the sum of the roots of the quadratic equation.
5
Apply Vieta's formulas to find the sum of the roots w1+w2w_1 + w_2.
The sum of the roots is w1+w2=−BA=−−(ln⁡2+ln⁡3)1=ln⁡2+ln⁡3w_1 + w_2 = -\frac{B}{A} = -\frac{-(\ln 2 + \ln 3)}{1} = \ln 2 + \ln 3. Using the logarithmic product rule, this simplifies to ln⁡(2⋅3)=ln⁡6\ln(2 \cdot 3) = \ln 6.
Vieta's formulas state that the sum of the roots of Aw2+Bw+C=0Aw^2 + Bw + C = 0 is −BA-\frac{B}{A}.
6
Calculate the product of the real values of xx.
x1⋅x2=eln⁡6=6x_1 \cdot x_2 = e^{\ln 6} = 6
Substituting the sum of the roots back into the exponent gives the final product.

Key Concept

Solving equations involving logarithmic properties, change of base, and relating quadratic roots to exponential functions
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Topics

Logarithmic and Exponential Expressions and Equations

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