Question

(2^х)^2×2^х+5=2×2^2×2^3×...×2^9×32

Answer 1

Ответ:15

Пошаговое объяснение:

Answer 2

To solve the equation (2^x)^2 * 2^x + 5 = 2 * 2^2 * 2^3 * ... * 2^9 * 32, you can follow these steps:

Step 1: Simplify the right-hand side of the equation. 2 * 2^2 * 2^3 * ... * 2^9 * 32 = 2^(1 + 2 + 3 + ... + 9) * 32

Now, find the sum of the exponents: 1 + 2 + 3 + ... + 9 is the sum of the first 9 positive integers, which is equal to (9 * 10) / 2 = 45.

So, the right-hand side simplifies to: 2^45 * 32 = 2^45 * 2^5 = 2^(45 + 5) = 2^50

Step 2: Substitute this simplified value back into the equation: (2^x)^2 * 2^x + 5 = 2^50

Step 3: Apply the properties of exponents. (2^(2x)) * 2^x + 5 = 2^50

Now, we have two terms with the same base (2), so we can combine them by adding their exponents: 2^(2x + x) + 5 = 2^50

Simplify the exponent on the left-hand side: 2^(3x) + 5 = 2^50

Step 4: Isolate the term with the exponent. 2^(3x) = 2^50 - 5

Step 5: Set the exponents equal to each other. 3x = 50 - 5

3x = 45

Step 6: Solve for x by dividing both sides by 3: x = 45 / 3 x = 15

So, the solution to the equation is x = 15.