4.

There are a few ways to do this question. Here's one way :

Change both expressions to in terms of ln first.


ln 3²⁰ / ln 2ˣ = ln 3²⁰²⁰ / ln 2ˣ⁺³

20ln 3 / xln 2 = 2020ln 3 / (x + 3)ln 2


Divide both sides by (20ln3 / ln2) ,


1/x = 101 / (x + 3)

Cross multiply,

x + 3 = 101x

100x = 3

x = 3/100


Since x = m/n, where m and n are relatively prime positive integers,

Then m = 3 and n = 100

m + n = 103

5.

7ˣ⁺⁷ = 8ˣ

7⁷ · 7ˣ = 8ˣ

7⁷ = 8ˣ / 7ˣ

7⁷ = (8/7)ˣ

logbase(8/7) 7⁷ = x


b = 8/7

By the way, the answer to your other deleted question about 100 + n² and 100 + (n+1)² :

①

If a common factor divides both numbers, then it divides their difference as well.

Why? Because both numbers are multiples of that factor. So their difference is also a multiple of it.

If two positive integers a and b (where a ≥ b) have an integer difference r , and a positive integer d divides them,

Then a = kd, b = jd, where k and j are also positive integers.

But a = b + r
kd = jd + r
r = kd - jd = (k - j)d

d also divides r.

So dₙ also divides 2n + 1 , which is the difference between 100 + n² and 100 + (n+1)²


The GCD of these two numbers, dₙ is also the GCD of 100 + n² and 2n + 1. There can't be any higher common factor (if not, it would divide 100 + (n+1)² as well)


2n + 1 and one of the numbers are always odd so 2 can never be the GCD.


②

So 100 + n² and 2n + 1 are both multiples of dₙ,
And, 100 + n² > 2n + 1 for all n

(Since 100 + n² - (2n + 1) = n² - 2n + 99
= (n - 2)² + 98

(n - 2)² ≥ 0 for all n so (n - 2)² + 98 > 0)


The biggest possible value of dₙ would be when dₙ = 2n + 1

In other words, when 100 + n² is a multiple of 2n + 1.


Doing polynomial long division,

(100 + n²)/(2n + 1)

= ¼(400 + 4n²)/(2n + 1)

= ¼(4n² - 1 + 401)/(2n + 1)

= ¼((2n+1)(2n-1) + 401)/(2n+1)

= ¼ [(2n - 1) + 401/(2n + 1)]


Now 2n - 1 is an odd integer and 401 is prime. We want the term inside the square brackets to be an integer such that it is a multiple of 2n + 1.

(If it is not even an integer then we don't even need to consider its divisibility by 4 (the ¼ you see above)


This means that 401/(2n + 1) must be an integer i.e the two integers will add up to give another integer


Since 401 is prime, then its only factors are 401 and 1. This means that 2n + 1 must = 401 such that 401/(2n + 1) is an integer.


So the largest dₙ = 401


Check :

2n + 1 = 401 means that :

n = 200
2n - 1 = 399
401/(2n + 1) = 1


So ¼ [(2n - 1) + 401/(2n + 1)]

= ¼ (399 + 1)
= ¼ (400)
= 100


401 × 100 = 40100 = 100 + 200²
40100 + 401 = 40501 = 100 + 201²

This also tells us one thing :


For any other n other than 200, 401/(2n + 1) will always be a fraction so the term in the square brackets will always be a mixed number.

So we can never get an integer multiple of (2n + 1) and this means that for all other n , 100 + n² and 2n + 1 are coprime.


So, this implies that 100 + n² and 100 + (n + 1)² are also coprime for all other n.


Something to note :

Why did we not consider the case where GCD could be 100 + n² itself?


Because, if that were the case, then the difference 2n + 1 would be a multiple of 100 + n² .

Eg. If 100 + (n+1)² = 5(100 + n²), then

2n + 1 = 4(100 + n²)

But this is not possible as we have already shown previously that 2n + 1 is always smaller than 100 + n²

So there would be no real solutions for 2n + 1 = k(100 + n²) where k is a positive integer.