Exactly how many digits are in 302627325(2^5300001+1)

we can ignore the "+1" since it adds a minuscule amount to the total.

log302627325(2^5300001)
= log 302627325 + 5300001 log2
= 1595467.76

so, I'd say there are 1,595,468 digits

To find the exact number of digits in a given number, we need to calculate the number first and then count the digits. In this case, we have the number 302,627,325 multiplied by the expression 2^5,300,001 + 1.

Let's break this down into steps to find the solution:

Step 1: Calculate the value of the expression 2^5,300,001.
To do this, we need to use an algorithm or a computer program that can handle large numbers. The result of this step will be a very large number.

Step 2: Add 1 to the result obtained in step 1.
This will give us another large number.

Step 3: Multiply 302,627,325 by the result obtained in step 2.
Again, we need to use an algorithm or a program that can handle large numbers to get the product.

Step 4: Count the number of digits in the final result obtained in step 3.
This will give us the answer to your question.

Due to the complexity of these calculations involving extremely large numbers, it is not feasible to perform them manually. Instead, you will need to use programming languages or calculators that support large number calculations.

Using programming languages like Python, you can write a script to calculate this number and count its digits. Here's an example of a Python script:

```python
import math

exponent = 5300001
base = 2
number = 302627325

# Calculate the value of the expression
result = pow(base, exponent) + 1

# Multiply the number by the result
final_result = number * result

# Count the number of digits
num_digits = int(math.log10(final_result)) + 1

print("The number of digits in the expression 302,627,325 * (2^5,300,001 + 1) is:", num_digits)
```

Running this script will give you the exact number of digits in the expression. Keep in mind that the result will be a very large number, and it might take some time for the calculation to complete.