Question

2log5(x^2-5x)/log5(x^2) ≤1

Answer 1

$\dfrac{2\log_5(x^2-5x)}{\log_5x^2}\leq1\\\\1)~(x^2-5x)>0~~~\Rightarrow~~~x(x-5)>0\\~~~~~x\in(-\infty;0)\cup(5;+\infty)\\\\2)~x^2>0~~~\Rightarrow~~~x\neq0\\~~~~~x\in(-\infty;0)\cup(0;+\infty)\\\\3)~\log_5x^2\neq0;~~~\Rightarrow~~~x^2\neq1~~~\Rightarrow~~~x\neq\pm1\\~~~~x\in(-\infty;-1)\cup(-1;1)\cup(1;+\infty)\\\\\boxed{\boldsymbol{D:~~x\in(-\infty;-1)\cup(-1;0)\cup(5;+\infty)}}$

$\dfrac{2\log_5(x^2-5x)}{\log_5x^2}\leq1~~~~\Leftrightarrow~~~~\dfrac{2\log_5(x^2-5x)}{\log_5x^2}-1\leq0\\\\\\\dfrac{\log_5(x^2-5x)^2-\log_5x^2}{\log_5x^2}\leq0\\\\\\\dfrac{\log_5\bigg(\dfrac{(x^2-5x)^2}{x^2}\bigg)}{\log_5x^2}\leq0~~~~\Leftrightarrow~~~~\dfrac{\log_5\bigg(\dfrac{x^2-5x}x\bigg)^2}{\log_5x^2}\leq0$

$\displaystyle\dfrac{\log_5(x-5)^2}{\log_5x^2}\leq0\\\\1)~\log_5(x-5)^2=0;~~~(x-5)^2=1\\\\~~~~~\left[\begin{array}{c}x-5=1\\x-5=-1\end{array}~~~\Rightarrow~~~\left[\begin{array}{c}x_1=6\\x_2=4\end{array}$

$2)~\log_5x^2\neq0;~~\Rightarrow~~x^2\neq1;~~\Rightarrow~~x_{3,4}\neq\pm1$

Метод интервалов  и область определения

$\underline{~~~~~~~~~~~~~~~~//////~~~~~~~~~~~~~~~~////////~~~~~~~~~}\\+++(-1)---(0)}..........(5)---[6]+++>x$

Ответ : x ∈ (-1; 0) ∪ (5;6]

Answer 2

To solve the inequality:

2log5(x^2-5x)/log5(x^2) ≤1

We can simplify the left side of the inequality using the rules of logarithms:

2log5(x^2-5x) - log5(x^2) ≤ 0

Now, we can apply the power rule of logarithms to get:

log5[(x^2-5x)^2/x^2] ≤ 0

Simplifying further:

log5[(x-5)^2/x^2] ≤ 0

Now, we can convert the inequality into exponential form:

5^0 ≤ (x-5)^2/x^2

Which simplifies to:

1 ≤ (x-5)^2/x^2

Multiplying both sides by x^2:

x^2 ≤ (x-5)^2

Expanding the right side:

x^2 ≤ x^2 - 10x + 25

Simplifying:

10x ≤ 25

x ≤ 2.5

Therefore, the solution to the inequality is:

x ≤ 2.5

Note that we must exclude any values of x that make the denominator of the logarithm, x^2, equal to 0 or negative. However, since x^2 is always positive for real values of x, we don't need to worry about this restriction.