Question

4^(x-1)+4^(x+2) ≤ 130 CРОЧНО!!!!

Answer 1

, берем  log по основанию  4 получаем х-1+х+2<=log(4)130
значение логарифма смотрим по таблице  или  пересчитываем из десятичного (㏒$_{4}$ a=lga\lg4- по-моему) и  решаем относительно  х

Answer 2

Solving the Inequality 4^(x-1) + 4^(x+2) ≤ 130

To solve the inequality 4^(x-1) + 4^(x+2) ≤ 130, we can use some algebraic manipulation and properties of exponents.

First, let's rewrite the inequality in terms of a single base:

4^(x-1) + 4^(x+2) ≤ 130

We know that 4^(x+2) can be rewritten as 4^x * 4^2 and 4^(x-1) can be rewritten as 4^x * 4^(-1).

So, the inequality becomes:

4^x * 4^(-1) + 4^x * 4^2 ≤ 130

Now, we can simplify this expression:

4^x * (1/4) + 4^x * 16 ≤ 130

Combining the terms:

(1/4) * 4^x + 16 * 4^x ≤ 130

Now, we can combine the terms with the same base:

(1/4 + 16) * 4^x ≤ 130

Solving for 4^x:

4^x ≤ 130 / (1/4 + 16)

Let's calculate the value of 130 / (1/4 + 16) to find the upper limit for 4^x.

130 / (1/4 + 16) = 130 / (65/4) = 130 * 4/65 = 520/65 = 8

So, we have:

4^x ≤ 8

Now, we need to find the value of x for which 4^x is less than or equal to 8. Taking the logarithm of both sides can help us find this value.

x ≤ log_4(8)

Using the property of logarithms that log_a(b) = log_c(b) / log_c(a), where a, b, c > 0:

x ≤ log(8) / log(4)

Calculating the value:

x ≤ 3/2

Therefore, the solution to the inequality 4^(x-1) + 4^(x+2) ≤ 130 is x ≤ 3/2.

This means that for any value of x less than or equal to 3/2, the inequality will hold true.

Conclusion

The solution to the inequality 4^(x-1) + 4^(x+2) ≤ 130 is x ≤ 3/2.