Hi Aditya!The given can be written as: {log(n - 1) + log(n^2 + n + 1)} / log (3*sqrt(2)) = 2-> log({n - 1} * {n^2 + n + 1}) = 2 log (3 * sqrt(2)) = log 18-> log(n^3 - 1) = log 18-> n^3 - 1 = 18   ->  n^3 = 19                So, n = (19) ^ (1/3)

loga(b)a^loga(b)=bso  (x-3)^logx+3(x^2 - x)< ( x+3)^1x^2 -x < x+3solve this equation you will get (-1,3)than in this equation we have one more function x^x - x this should be > 1 you will get 0 and 1another equation  x+3 not equal to 1 and should be >0 than take the intersection of all domain basically we try to find the domain of all possible function that is present in equation

Logarithms IMG_20240712_213219.jpg 54.39 KB

Logarithms IMG_20240712_213219.jpg 54.39 KB

Logarithms How to solve this question?Screenshot_2024-07-26-10-58-35-59_f9ee0578fe1cc94de7482bd41accb329.jpg 213.43 KB