log_2(4) * log_2(5) * log_2(6) * ... * log_2(n + 1) = 4
To solve for n we can use the logarithmic properties and solve for n. The solution to the problem is n = 16².
Here's how to solve the equation:
log_2(4) * log_2(5) * log_2(6) * ... * log_2(n + 1) = 4
First, we can use the logarithmic property that states that the product of logarithms is equal to the logarithm of the product. This gives us:
log_2(4 * 5 * 6 * ... * (n + 1)) = 4
Next, we can simplify the product inside the logarithm:
log_2(4 * 5 * 6 * ... * (n + 1)) = 4
log_2(4 * 5 * 6 * ... * n * (n + 1)) = 4
log_2(4) + log_2(5) + log_2(6) + ... + log_2(n) + log_2(n + 1) = 4
```
Since we know that `log_2(4) = 2`, `log_2(5) = 2.32`, and so on, we can substitute these values into the equation:
2 + 2.32 + 2.58 + ... + log_2(n) + log_2(n + 1) = 4
We can then simplify this equation by subtracting `2` from both sides:
0.32 + 0.58 + ... + log_2(n) + log_2(n + 1) = 2
Next, we can use the logarithmic property that states that the sum of logarithms is equal to the logarithm of the product. This gives us:
log_2((n)(n+1)/3!) = 1.68
We can then solve for `n` by exponentiating both sides of the equation:
(n)(n+1)/3! = 3.39
(n)(n+1) = (3.39)(6)
n^2 + n - (3.39)(6) = 0
Using the quadratic formula, we get:
n = (-1 ± sqrt(1 + (4)(3.39)(6))) / (2)
n ≈ -0.17 or n ≈ 16
Since n must be a positive integer, we can discard the negative solution and conclude that n ≈ 16¹.
Therefore, the solution to your problem is n = 16