When is exponentiating both sides of equation an equivalent operation? I have the following inequation: $\sqrt{t^{2}-t-12}<7-t$. Can I just set both sides of the inequation to the power of two, or is there any condition under which exponentiating is an equivalent operation?
Thanks
 A: This is not an equation; it's an inequality.
However, you have that $\exp\text{LHS}<\exp\text{RHS}$ because the exponential function is monotonic.
Oh, it seems the operation you want to perform is not exponentiation, but empowerment -- raising to a power. You may always do this if the power you want to raise both sides to is odd or a fraction with odd denominator since such functions are monotonic. However, when they are even, or have even denominator, you have to make sure that both sides are nonnegative. Otherwise, you need to simultaneously reverse the inequality, too.
Thus, you may square both sides without changing the $<$ to $>$ only whenever $7-t\ge 0.$ Of course, $\text{LHS},$ being a square root, is always nonnegative whenever it is real.
You have used many terms wrongly here, so let me correct the last: equivalent operation. Well, it seems like what you mean is order-preserving operation. What we have are equivalence relations, which are quite different things from operations that respect linear orders.
A: Hints:


*

*You have to specify the domain of validity of the inequation. Here the condition is $t^2-t-12\ge 0$. As the roots of this quadratic polynomial are $-3$ and $4$, you obtain $\;(-\infty,-3]\cup[4,+\infty)$.

*In the domain of validity, you have the equivalence
$$\sqrt{A}<B\iff A<B^2 \; \color{red}{\text{ and }B\ge 0}.$$
A: It is valid to exponentiate if you preserve the inequality, that is, if $$a<b\iff a^c<b^c$$
We would have a problem if we were to do this as follows: $$-2<1\iff (-2)^2<1^2$$
In your case there is no problem, since the squareroot is defined to be positive, we only have to make sure that it stays that way. We have to look at $t:t^2-t-12\geq 0$, that is, $t\in (-\infty,-3]\cup [4,\infty)$, and also $t< 7$ (otherwise we would have $\sqrt{t^2-t-12}<0$, which is a problem). This forces us to have $t\in (-\infty,-3]\cup [4,7)$.
Given these conditions we know that $$\sqrt{t^2-t-12}<7-t\iff t^2-t-12<(7-t)^2$$ This simplifies to $$t<\frac{61}{13}$$
A: It must be $$t^2-t-12\geq 0$$ and $$7-t>0$$, then we get by raising to the power two:
$$t^2-t-12<49+t^2-14t$$ nand we get
$$t<\frac{61}{13}$$
Finally we get
$$t\le -2\sqrt{3}$$
