Question

(a) $\log_{x+1}(x^2+x-6)=4$

Answer 1

No real solution.

Answer 2

For the logarithm to be defined, the base $x+1$ must be positive and not equal to 1, so $x > -1$ and $x \neq 0$. The argument $x^2+x-6$ must be positive, which means $(x+3)(x-2) > 0$, so $x < -3$ or $x > 2$. Combining these conditions, the domain is $x > 2$. Converting the logarithmic equation to exponential form: $(x+1)^4 = x^2+x-6$. Expanding the left side: $x^4+4x^3+6x^2+4x+1 = x^2+x-6$. Rearranging into a polynomial equation: $x^4+4x^3+5x^2+3x+7 = 0$. For $x > 2$, all terms $x^4$, $4x^3$, $5x^2$, $3x$, and $7$ are positive. Thus, their sum is always positive and cannot be equal to zero. Therefore, there are no real solutions for $x > 2$.