Question

(a)log$_{x+1}$(x$^{2}$+x-6)=4

Answer 1

No solution

Answer 2

The domain of the equation requires $x > 2$. Converting the logarithmic equation to exponential form leads to $(x+1)^4 = x^2+x-6$. Substituting $u=x+1$, we obtain the quartic equation $u^4 - u^2 + u + 6 = 0$. The domain $x>2$ implies $u>3$. The function $f(u) = u^4 - u^2 + u + 6$ is strictly increasing for $u>3$, and $f(3)=81$. Thus, for $u>3$, $f(u)>81$, meaning there are no real solutions to $f(u)=0$ in the required range.