Question

4√a^x-1=3√a^x-2 3√29+√(27^2-22^2•5)

Answer 1

1. 4√a^x - 1 = 3√a^x - 2. Перенесем все с х в одну сторону, а без - в другую
√a^x = -1, но
√a^x > 0, поэтому решений нет.
2. 3√29 + √(27^2 - 22^2*5) = 3√29 + √(729 - 484*5) = 3√29 + √(729 - 2420) = 3√29 + √-1691

Answer 2

Let's break down the given expressions step by step.

Expression 1: 4√(a^(x-1)) = 3√(a^(x-2))

To solve this equation, we need to isolate the variable x. Let's begin by cubing both sides of the equation to eliminate the cube roots:

(4√(a^(x-1)))^3 = (3√(a^(x-2)))^3

Simplifying further:

(4^3) * (√(a^(x-1)))^3 = (3^3) * (√(a^(x-2)))^3

64 * (a^(x-1)) = 27 * (a^(x-2))

Now, let's solve for x by simplifying the equation:

64 * a^(x-1) = 27 * a^(x-2)

Dividing both sides by a^(x-2):

64 * (a^(x-1 - (x-2))) = 27

64 * (a^1) = 27

64 * a = 27

x = logₐ(27/64)

Expression 2: 3√(29 + √(27^2 - 22^2 * 5))

First, let's simplify the expression inside the square root:

27^2 - 22^2 * 5

= 729 - 484 * 5

= 729 - 2420

= -1691

Since the value inside the square root is negative, we can't take the square root directly. Therefore, the expression is not defined for real numbers.

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