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Math  /  Algebra

Question(15 points) Solve the equation below, finding all real solutions. Write your final answer(s) in the box provided. log⁡6(2x−1)+log⁡6(x)=1\log _{6}(2 x-1)+\log _{6}(x)=1

Studdy Solution

STEP 1

1. The equation involves logarithms with the same base, which allows us to use logarithmic properties to simplify.
2. The domain of the logarithmic functions requires that the arguments 2x−12x - 1 and xx must be positive.
3. We will use properties of logarithms to combine and simplify the equation.

STEP 2

1. Use the properties of logarithms to combine the logarithmic terms.
2. Convert the logarithmic equation into an exponential equation.
3. Solve the resulting polynomial equation.
4. Check the solutions against the domain restrictions.

STEP 3

Use the property of logarithms that states log⁡b(A)+log⁡b(B)=log⁡b(A⋅B)\log_b(A) + \log_b(B) = \log_b(A \cdot B) to combine the terms:
log⁡6((2x−1)⋅x)=1\log_6((2x - 1) \cdot x) = 1

STEP 4

Convert the logarithmic equation to an exponential equation. Recall that if log⁡b(A)=C\log_b(A) = C, then A=bCA = b^C:
(2x−1)⋅x=61(2x - 1) \cdot x = 6^1 2x2−x=62x^2 - x = 6

STEP 5

Rearrange the equation to form a standard quadratic equation:
2x2−x−6=02x^2 - x - 6 = 0
Solve this quadratic equation using the quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=2a = 2, b=−1b = -1, and c=−6c = -6:
x=−(−1)±(−1)2−4⋅2⋅(−6)2⋅2x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-6)}}{2 \cdot 2} x=1±1+484x = \frac{1 \pm \sqrt{1 + 48}}{4} x=1±494x = \frac{1 \pm \sqrt{49}}{4} x=1±74x = \frac{1 \pm 7}{4}
This gives us two potential solutions:
x=84=2x = \frac{8}{4} = 2 x=−64=−32x = \frac{-6}{4} = -\frac{3}{2}

STEP 6

Check the solutions against the domain restrictions. The arguments of the logarithms must be positive:
For x=2x = 2: - 2x−1=2(2)−1=3>02x - 1 = 2(2) - 1 = 3 > 0 - x=2>0x = 2 > 0
For x=−32x = -\frac{3}{2}: - 2x−1=2(−32)−1=−3−1=−4<02x - 1 = 2(-\frac{3}{2}) - 1 = -3 - 1 = -4 < 0 - x=−32<0x = -\frac{3}{2} < 0
The solution x=−32x = -\frac{3}{2} is not valid because it does not satisfy the domain restrictions.
The valid real solution is:
2 \boxed{2}

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