Question

2 ㏒ ₉ ( 2 - x ) - ㏒ ₃ x = 0

Answer 1

Ответ:

Пошаговое объяснение:

$2log_9(2-x)-log_3x=0\\2log_{3^2}(2-x)=log_3x\\ODZ: x>0\\2-x>0\\-x>-2\\x$

⇒x∈(0;2)

$log_3(2-x)=log_3x\\2-x=x\\2x=2\\x=1$

Answer 2

The equation you've provided contains logarithmic terms. To solve for the variable x, we need to manipulate the equation to isolate x on one side. The equation is:

$2\log_9(2 - x) - \log_3 x = 0$

Let's work through the steps to solve for x:

Step 1: Combine the logarithmic terms using logarithmic properties. $\log_9((2 - x)^2) - \log_3 x = 0$

Step 2: Use the property $\log_a b - \log_a c = \log_a \frac{b}{c}$. $\log_9\left(\frac{(2 - x)^2}{x}\right) = 0$

Step 3: Remove the logarithm by exponentiating both sides with base 9. $\frac{(2 - x)^2}{x} = 9^0$ $\frac{(2 - x)^2}{x} = 1$

Step 4: Cross-multiply to eliminate fractions. $(2 - x)^2 = x$

Step 5: Expand the square on the left side. $4 - 4x + x^2 = x$

Step 6: Move all terms to one side to form a quadratic equation. $x^2 - 5x + 4 = 0$

Step 7: Factor the quadratic equation. $(x - 4)(x - 1) = 0$

Now, we have two possible solutions:

1. $x - 4 = 0 \Rightarrow x = 4$
2. $x - 1 = 0 \Rightarrow x = 1$

So, the solutions for the given equation are $x = 4$ and $x = 1$.