Question

125^x > 25^(x^2-1) папаките поже

Answer 1

Вроде бы все Легко,как это можно не знать ?

Answer 2

To solve the inequality $125^x > 25^{x^2-1}$, we need to compare the bases and exponents on both sides of the inequality. Let's simplify the inequality step by step:

Step 1: Write the bases with the same exponent (common base) on both sides of the inequality.

Recall that $125$ can be written as $5^3$, and $25$ can be written as $5^2$. So, our inequality becomes:

$(5^3)^x > (5^2)^{x^2-1}$

Step 2: Simplify the exponents.

$(5^{3x}) > (5^{2(x^2-1)})$

Step 3: Apply the property of exponents, which states that for any positive $a$, $b$, and $c$:

$(a^b)^c = a^{b \cdot c}$

Using this property, we can simplify further:

$5^{3x} > 5^{2x^2 - 2}$

Step 4: Since the bases are the same (both are $5$), we can compare the exponents directly:

$3x > 2x^2 - 2$

Step 5: Move all terms to one side to set the inequality to zero:

$2x^2 - 3x + 2 < 0$

Now, we have a quadratic inequality. To find the solution, we need to determine the values of $x$ that make the inequality true. One way to do this is by finding the critical points and testing intervals.

Step 6: Find the critical points by setting the expression inside the inequality to zero:

$2x^2 - 3x + 2 = 0$

However, the discriminant ($b^2 - 4ac$) of this quadratic equation is negative, indicating that there are no real roots. Since the coefficient of the $x^2$ term is positive ($2 > 0$), the parabola opens upward, meaning it's positive for all $x$.

Step 7: Determine the sign of the quadratic expression for different intervals:

Since the quadratic expression is positive for all $x$, the inequality $2x^2 - 3x + 2 < 0$ has no solutions in the real number domain.

In conclusion, the original inequality $125^x > 25^{x^2-1}$ has no real solutions for $x$.