Question

Log2^2(2-x) +5log2(2-x)=6

Answer 1

Ответ:

Делаем замену t = log2(x), получаем:

t^2 - 5t + 6 = 0; Решаем квадратное уравнение: t1 = 2; t2 = 3;

Значит, log2(x) = 2 и log2(x) = 3, решаем дальше:

log2(x) = 2*log2(2); log2(x) = log2(2^2); log2(x) = log2(4); x = 4;

Аналогично для log2(x) = 3; log2(x) = 3*log2(2); log2(x) = log(2^3); x = 8

Объяснение: вот

Answer 2

To solve the equation $\log_2^2(2-x) + 5\log_2(2-x) = 6$, you can use substitution and some logarithmic properties. Let's break it down step by step:

1. Start with the given equation: $\log_2^2(2-x) + 5\log_2(2-x) = 6$

2. Rewrite the first term as a square of a logarithm: $(\log_2(2-x))^2 + 5\log_2(2-x) = 6$

3. Let $y = \log_2(2-x)$. Now, we have a quadratic equation in terms of $y$: $y^2 + 5y = 6$

4. Rearrange the equation: $y^2 + 5y - 6 = 0$

5. Factor the quadratic equation: $(y + 6)(y - 1) = 0$

6. Solve for $y$ by setting each factor equal to zero: $y + 6 = 0$ or $y - 1 = 0$

7. Solve for $y$ in each equation: $y = -6$ or $y = 1$

8. Recall that $y = \log_2(2-x)$. We can now solve for $x$ in each case:

   Case 1: $y = -6$ $\log_2(2-x) = -6$

   Using the definition of logarithms, we get: $2-x = 2^{-6}$

   Simplify: $2-x = \frac{1}{64}$

   Now, solve for $x$: $x = 2 - \frac{1}{64}$ $x = \frac{127}{64}$

   Case 2: $y = 1$ $\log_2(2-x) = 1$

   Using the definition of logarithms again: $2-x = 2^1$

   Simplify: $2-x = 2$

   Now, solve for $x$: $x = 2 - 2$ $x = 0$

So, the solutions to the equation are $x = \frac{127}{64}$ and $x = 0$.