For f(x) = ((3x+7)^8)((4x-5)^3) , find f'(x) and use this answer to find the value(s) of x at which the graph of f(x) has a horizontal tangent line I know that $f'(x) = (24(3+7)^7)*(4x-5)^3+((3x+7)^8)*12(4x-5)^2$, but is there any easier way to find the horizontal tangent line without expanding the terms using pascal's triangle and solving for x by setting f'(x) equal to 0?

Hint:
f(x) = $(3x+7)^8 * (4x-5)^3$
log(f(x)) =  $8log(3x+7) + 3log(4x-5)$
Now take the derivative and get the required answer. It's a one-two step process.
