Question

С ОДЗ 1)log 1/3 (log5 x) ) >= 0

Answer 1

log 1/3 (log5 x) ) >= 0   5x>0   x>0
log 3 log5x≤0   log3 5x≤3⁰=1   5x≤3   x≤3.5
x∈(0;3.5]

Answer 2

To solve the inequality $\log_{\frac{1}{3}}(\log_5 x) \geq 0$, we'll break it down into smaller steps:

Step 1: Solve $\log_{\frac{1}{3}}(\log_5 x) = 0$. Step 2: Determine the critical points where $\log_{\frac{1}{3}}(\log_5 x)$ may change sign. Step 3: Check the sign of $\log_{\frac{1}{3}}(\log_5 x)$ in each interval defined by the critical points.

Let's go through each step:

Step 1: Solve $\log_{\frac{1}{3}}(\log_5 x) = 0$: To find the points where the logarithm is equal to zero, we set the argument equal to 1 (since $\log_b 1 = 0$ for any positive base $b$): $\log_5 x = 1$ Now, rewrite this equation in exponential form: $5^1 = x$ $x = 5$

Step 2: Determine the critical points where $\log_{\frac{1}{3}}(\log_5 x)$ may change sign: The expression inside the outer logarithm, $\log_5 x$, must be positive since $\log_{\frac{1}{3}}(y)$ is only defined for positive values of $y$. Thus, we need to solve the inequality: $\log_5 x > 0$ To solve this, recall that $\log_b(y) > 0$ when $y > 1$, so we have: $\log_5 x > 0 \quad \Rightarrow \quad x > 1$

Step 3: Check the sign of $\log_{\frac{1}{3}}(\log_5 x)$ in each interval defined by the critical points: We have one critical point at $x = 5$. Now, we can check the sign of $\log_{\frac{1}{3}}(\log_5 x)$ in the intervals $(1, 5)$ and $(5, \infty)$.

Interval $(1, 5)$: Choose $x = 2$ (a value within the interval). We have: $\log_{\frac{1}{3}}(\log_5 2) > 0$ $\log_{\frac{1}{3}}(> 0) > 0$ Since the logarithm of a positive number with a base between 0 and 1 is positive, this is true.

Interval $(5, \infty)$: Choose $x = 10$ (a value within the interval). We have: $\log_{\frac{1}{3}}(\log_5 10) > 0$ $\log_{\frac{1}{3}}(> 0) > 0$ Again, this is true.

So, the solution to the inequality $\log_{\frac{1}{3}}(\log_5 x) \geq 0$ is $x \in (1, \infty)$.