Prove this with the definition of the Θ notation

$ (3 n + 13) (7 n + 2) left ( log left (1024 n ^ {2} + 100 right) right) in

Theta left (n ^ {2} log n right) $

I also found a useful example in Stack exchange:

Big theta proof for the polynomial function

But my example has $ log $ Function, these are my steps:

apparently

begin {equation}

g (n) = n ^ {2} log n, f (n) = (3 n + 13) (7 n + 2) left ( log left (1024 n ^ {2} + 100 right) right)

end {equation}

Then we can get to the denomination of big-theta

begin {equation}

0 leqslant c_ {1} n ^ {2} log n leqslant (3 n + 13) (7 n + 2) left ( log left (1024 n ^ {2} + 100 right) right ) leqslant c_ {2} n ^ {2} log n

end {equation}

Divide the inequality by the largest n-term of the order we get

begin {equation}

0 leqslant c_ {1} leqslant left (21+ frac {97} {n} + frac {26} {n ^ {2}} right) log _ {n} left (1024 n ^ {2} +100 right) leqslant c_ {2}

end {equation}

(I see that $ n neq 1 $I think it's right, so I choose $ n geqslant $ 2)

by calculate the limit

begin {equation}

lim _ {n rightarrow infty} left (21+ frac {97} {n} + frac {26} {n ^ {2}} right) log _ {n} left (1024 n ^ {2} +100 right) = 42

end {equation}

we know $ c_2 = 42 $ when $ n geqslant $ 2, then I should choose a constant that is less than 42, then it should meet the LHS.

I choose $ c_1 = 41 $, To the $ n $I choose too $ 2 $ that can satisfy LHS.

So the constants that prove

$ (3 n + 13) (7 n + 2) left ( log left (1024 n ^ {2} + 100 right) right) in

Theta left (n ^ {2} log n right) $

are $ c_1 = 41, c_2 = 42, n geqslant $ 2

My steps or answers are right or wrong? Please explain or correct my mistake.