Question

Помогите пожалуйста! 1) найти наименьшее целое x удовлетворяющее неравенству 10 в степени 2x/7 >0,1

Answer 1

10^2x/7>0,1;
10^2x/7>10^-1;
2x/7>-1;
2x>-7;
x>-3,5.
Наименьшее целое x, удовлетворяющее неравенству - -3.
ОТВЕТ: -3.

Answer 2

I can help you with your math problem.

To find the smallest integer x that satisfies the inequality $$10^{2x/7}>0.1$$, you need to do the following steps:

- Take the logarithm of both sides with base 10: $$\log_{10}(10^{2x/7})>\log_{10}(0.1)$$ - Simplify the left side using the property of logarithms: $$\frac{2x}{7}>\log_{10}(0.1)$$ - Evaluate the right side using a calculator: $$\frac{2x}{7}>-1$$ - Multiply both sides by 7 and divide by 2: $$x>-3.5$$ - Since x must be an integer, the smallest possible value is -3.

Therefore, the answer is -3. You can check your answer by plugging it into the original inequality and see if it is true.

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