Question

1/2log2(х-4)+1/2log2(2х-1)=log2(3)

Answer 1

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Answer 2

Решается по свойствам логарифмов

Answer 3

To solve the equation $\frac{1}{2} \log_2(x-4) + \frac{1}{2} \log_2(2x-1) = \log_2(3)$, you can use properties of logarithms and algebraic techniques. Here's the step-by-step solution:

1. Use the properties of logarithms to combine the two logarithmic terms on the left-hand side:

   $\frac{1}{2} \log_2(x-4) + \frac{1}{2} \log_2(2x-1) = \log_2(3)$

   $\log_2((x-4)^\frac{1}{2} \cdot (2x-1)^\frac{1}{2}) = \log_2(3)$

2. Remove the logarithm by equating the expressions inside the logarithms:

   $(x-4)^\frac{1}{2} \cdot (2x-1)^\frac{1}{2} = 3$

3. Square both sides of the equation to get rid of the square roots:

   $((x-4)^\frac{1}{2} \cdot (2x-1)^\frac{1}{2})^2 = 3^2$

   $(x-4) \cdot (2x-1) = 9$

4. Expand and simplify the equation:

   $2x^2 - x - 8x + 4 = 9$

   $2x^2 - 9x - 5 = 0$

5. To solve this quadratic equation, you can use the quadratic formula:

   $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

   In this case, $a = 2$, $b = -9$, and $c = -5$.

   $x = \frac{9 \pm \sqrt{(-9)^2 - 4(2)(-5)}}{2(2)}$

   $x = \frac{9 \pm \sqrt{81 + 40}}{4}$

   $x = \frac{9 \pm \sqrt{121}}{4}$

   $x = \frac{9 \pm 11}{4}$

Now, we have two possible solutions:

1. $x = \frac{9 + 11}{4} = \frac{20}{4} = 5$

2. $x = \frac{9 - 11}{4} = \frac{-2}{4} = -0.5$

However, it's important to check these solutions in the original equation because the logarithm of a non-positive number is undefined. Since the original equation contains logarithmic terms, we need to ensure that both $x = 5$ and $x = -0.5$ are valid solutions.

Checking $x = 5$: $\frac{1}{2} \log_2(5-4) + \frac{1}{2} \log_2(2(5)-1) = \log_2(3)$