CPhill

Very nice, ND.....you missed your calling.....you should have been a graphic artist!!

'c" is the correct answer

We have

1/x ≤ x         multiply both sides by x   .....note that we're assuming that x ≠ 0

1 ≤ x²          subtract x² from both sides

1- x² ≤ 0        I like to turn the inequaliy into an equality and factor ....this gives us some "interval" points to work with!!

(1 - x) (1 + x) = 0       Then x = ± 1

Our answer comes from two of four intervals (-∞, -1], [-1, 0), (0, 1] or [1, ∞)

If you pick a point in each interval, you'll find that [-1, 0) and [1, ∞)  are the intervals that "work' in the original inequality.......

Here's the code......

\begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array}

But...I'm still in Sydney !!  ....I just took the "scenic route"

We don't even need a calculator for this one. It basically says....what power do we raise 81 to, to get 3??

In exponential terms, we have

81^x = 3¹  

(3⁴)^x = 3¹

3^4x = 3¹

And solving for the exponents....we have

4x = 1

x = 1/4  = .25

And that's the answer

WOW...that's a pretty cool image!!!...I might even consider that as a new avatar - it's time to change mine, anyway......thanx......

$$\begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array}$$

What a genius I am....!!!!   You know....this copying and pasting code stuff is pretty brain-draining ....!!!!

Actually, Anonymous, it's sometimes hard to determine what is being asked !!! The lack of parentheses, brackets, etc., is sometimes troublesome.

I'll give Aziz the benefit of the doubt, here.  Besides, he may have answered the exact question that was being asked.....

Aw, heck....!!! You mean Aziz and I can't play, too???

I'm taking my marbles and going home........

Just add them together......