Question

(log0,5 x)*(1+log0,5 8/x)+3=0

Answer 1

(log0,5 x)*(1+log0,5 8/x)+3=0                  ОДЗ  x>0
(log0,5 x)*(1+log0,5 8-log0,5x)+3=0
(log0,5 x)*(1-3-log0,5 x)+3=0
(log0,5 x)*(-2-log0,5 x)+3=0
(log0,5 x)²+2(log0,5 x)-3=0
(log0,5 x)=a
a²+2a-3=0
a1+a2=-2 U a1*a2=-3
a1=-3⇒(log0,5 x)=-3⇒x=8
a2=1⇒(log0,5 x)=1⇒x=0,5

Answer 2

To solve the equation $(\log_{0.5}x) \cdot (1 + \log_{0.5}\frac{8}{x}) + 3 = 0$, you can follow these steps:

1. Start by subtracting 3 from both sides of the equation: $(\log_{0.5}x) \cdot (1 + \log_{0.5}\frac{8}{x}) = -3$

2. Now, you can use a substitution to simplify the equation. Let $u = \log_{0.5}x$. This means $x = 0.5^u$. Also, let $v = \log_{0.5}\frac{8}{x}$. This means $\frac{8}{x} = 0.5^v$.

   The equation now becomes: $u(1 + v) = -3$

3. Solve for $u$: $u = \frac{-3}{1+v}$

4. Substitute the expressions for $u$ and $v$ back in terms of $x$: $\log_{0.5}x = \frac{-3}{1+\log_{0.5}\frac{8}{x}}$

5. Now, let's isolate $\log_{0.5}\frac{8}{x}$ on the right side: $\log_{0.5}\frac{8}{x} = \frac{-3}{\log_{0.5}x} - 1$

6. To simplify further, let's use a change of base formula to switch the base of the logarithm from 0.5 to a common base, like 10 or e. Let's use base 10 in this case: $\frac{\log_{10}\frac{8}{x}}{\log_{10}0.5} = \frac{-3}{\log_{10}x} - 1$

7. Since $\log_{10}0.5$ is a constant, let $k = \frac{\log_{10}\frac{8}{x}}{\log_{10}0.5}$: $k = \frac{-3}{\log_{10}x} - 1$

8. Now, isolate $\log_{10}x$ on the right side: $\log_{10}x = \frac{-3}{k + 1}$

9. Finally, solve for $x$: $x = 10^{\frac{-3}{k + 1}}$

So, the solution for the equation $(\log_{0.5}x) \cdot (1 + \log_{0.5}\frac{8}{x}) + 3 = 0$ is $x = 10^{\frac{-3}{k + 1}}$, where $k = \frac{\log_{10}\frac{8}{x}}{\log_{10}0.5}$.