Solve for x : $ \ln (x^2-x-1) > 0 $

Question

$$ \text{Solve for x } $$ $$ \ln (x^2-x-1) > 0 $$

Here is how I proceded

So we have to basically solve these 2 equations

By the log rules $$ \ x^2-x-1>1 $$

And since domain of log is $(0,\infty)$

$$ x^2-x-1>0 $$

Upon solving these two equations , the solution that I got was as follows

$$ x\in \left({-\infty},\frac{-\sqrt{5}+1}{2}\right) \cup \biggl(2,\infty\biggr) $$

But the answer in textbook and even on online calculator is

$$ x\in \biggl({-\infty},1\biggr) \cup \biggl(2,\infty\biggr) $$

Is there any mistake in how I solved the question ?

Thanks in advance

Answer 1

You need $x^{2}-x-1>1$ or $x^{2}-x-2>0$ or $(x+1)(x-2) >0$. This gives $ x>2$ or $x <-1$.

$x^{2}-x-1>1$ automatically implies $x^{2}-x-1>0$ and there is no need to worry about domain of $\log$.

Answer 2

No need to make reference to the domain of $\ln$. You already had from the beginning that $$x^2-x-1>1.$$ Any value of $x$ that satisfies this inequality will satisfy the original one. So in fact, we have an equivalence.

To finish the problem, we just notice that $$\left(x-2\right)\left(x+1\right)> 0\Leftrightarrow $$ $$ x\in \left({-\infty},-1\right) \cup \left(2, \infty\right). $$

Answer 3

$$\ln (x^2-x-1)>0$$ So $x^2-x-1>0 \implies x < \frac{1-\sqrt{5}}{2}$ or $x > \frac{1+\sqrt{5}}{2}~~~(1)$. Next, $x^2-x-1>1 \implies x <-1$ or $x>2. ~~~(2).$ finally the overlap of (1) and (2) gives the answer as $x\in (-\infty, -1) \cup (2, \infty).$

Answer 4

Solution: $$\ln(x^2-x-1)>0$$ As log is defined for values $> 0$ , $x^2-x-1$ should be $> 0$, And as $e^x$ is an increasing function, applying $e^{f(x)}$ on both sides, gives us : $x^2-x-1>1$

Remember, we need a solution which satisfies the 1st inequality and 2nd inequality not or. On solving, you would get x belongs to $$\left(-\infty,\frac{1-\sqrt5}{2}\right) \cup \left(\frac{1+\sqrt5}{2}, +\infty\right) $$ for 1st inequality.

And for second one , $$(-\infty,-1) \cup (2,+\infty)$$

We need the solution which satisfies the both, in set theory, and refers to intersection of sets, so our final answer is,$ (-\infty,-1) \cup (2,\infty)$

or instead by a clear examination you would notice that just by solving the 2nd inequality we would get an answer for 1st one as well, as $1>0.$

The part where you have done wrong is, either you misplaced $-1$ with $+1$ and took intersection of sets or, you have considered union of sets instead of intersection.

Hope it helps, sorry for the typos(if any).