Question

1/2 log2 (x-2)^2+1/3 log2(x-4)^3=3​

Answer 1

log(a^k) b^b = b/k log(a) b

log a + log b = log ab

1/2 log2 (x-2)^2+1/3 log2(x-4)^3=3​

одз x ≠ 2   x>4

1/2 *2  log2 (x-2)+1/3 *3  log2(x-4) = 3​

log2 (x-2) + log2(x-4) = 3​

log2 (x-2)(x-4) = 3​

(x-2)(x-4) = 8

​x^2 - 2x - 4x + 8 = 8

​x(x - 6) = 0

x = 0 не проходит по одз

x = 6

Answer 2

Ответ:

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x ∈ (4; +∞)

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$x^2 - 6x + 8 = 8\\x (x-6) = 0\\\\\ \left[\begin{array}{ccc}x = 0\\x = 6\\x>4 \right\end{array}\right$

Ответ: 6.

Answer 3

To solve the equation $\frac{1}{2} \log_2{(x-2)^2} + \frac{1}{3} \log_2{(x-4)^3} = 3$, you can use logarithm properties to simplify and solve for $x$. Here are the steps:

1. Start by simplifying the equation using logarithm properties. Recall that $\log_a{(x^y)} = y\log_a{x}$:

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

Rewrite the exponents as coefficients:

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

Simplify further:

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

2. Now, apply the properties of logarithms. When you add logarithms with the same base, you can multiply the arguments:

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

3. Remove the logarithm by exponentiating both sides. In this case, we'll use base 2:

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

4. Simplify the right side:

$$(x-2)(x-4) = 8$$

5. Expand the left side by multiplying the terms:

$$x^2 - 6x + 8 = 8$$

6. Subtract 8 from both sides:

$$x^2 - 6x = 0$$

7. Factor the equation:

$$x(x - 6) = 0$$

8. Set each factor equal to zero and solve for $x$:

a) $x = 0$

b) $x - 6 = 0 \Rightarrow x = 6$

So, the solutions to the equation are $x = 0$ and $x = 6$.