Question

Given 3 positive intergers a, b, and c. Knowing a < b < c. If 42ª + 2² +2²=2336, what is the value of a?

Answer 1

The value of 'a' in the equation 42^a + 2^2 + 2^2 = 2336, given that a, b, and c are positive integers and a < b < c, is 5.

Step-by-step explanation:

To find the value of a, we can use the given equation and substitute the values of b and c. The equation is 42^a + 2^2 + 2^2 = 2336. By simplifying this equation, we get 42^a + 8 = 2336. Subtracting 8 from both sides, we have 42^a = 2328.

We are asked to solve the following equation: 42a + 22 + 22 = 2336. Given that a, b, and c are positive integers and a < b < c, we will solve for the value of a.

First, combine the terms with the same base:

• 42a + 4 + 4 = 2336
• 42a = 2328
• 42a = 25 × 73

Since 42a must be a power of 2 for the equality to hold and 73 is not a power of 2, the only option is 25. Hence, a must be 5.

Answer 2

Answer: We are given that:

42^a + 2^2 + 2^2 = 2336

Simplifying the expression on the left-hand side, we get:

42^a + 4 = 2336

Subtracting 4 from both sides, we get:

42^a = 2332

Taking the logarithm of both sides with base 42, we get:

a = log₄₂₂₋₃₂ ≈ 1.417

Since a is a positive integer, the closest integer to 1.417 is 1. Therefore, the value of a is 1.

Step-by-step explanation: