For how many real numbers does ?
0,1,2,3 or infinitely many?
Hmmm....
Oh yea, no graphing calculator for this question.
hmmm, the easiest way to see the answer is through graphs. of course, i didn't use a graphing calculator either knowing what these graphs look like in your head makes the answer obvious. the other alternative i can think of is too complicated (using Newton's method and a whole lot of speculation)
the answer is 1 by the way ....note that
well, for positive x-values, e^x is always bigger than ln(x), so there is no intersection there. but for negative x values, e^x keeps decreasing tending to zero, while ln(-x) keeps increasing, tending to infinity. the shapes of the graphs make it obvious they will cross each other only once, as one is strictly increasing and the other is strictly decreasing
see below. of course, you should know how this looks in your head, no graphing tool necessary
Hello, JOhkonut!
You know that looks like, right?For how many real numbers does ?Code:| * | | * | * | * * * | * | - - - - - - - + - - - - - - - |
And you know what looks like.
. . (It's the inverse of the exponential function.)
Then is the same graph
. . but with a reflection over the y-axis.Code:| * | * * | * - - - - -*- - + - -*- - - - - * | * * | * | *|* |
Now place them on the same graph and count the intersections.
Darn . . . too slow again!
.
Hello !
One way to do it, without graph, is to study the function . Now, be patient and read carefully
IF X<0
Which is not possible, because and
Therefore, and f is strictly increasing when x<0.
(x=0 is a vertical asymptote)
-----------------------------------------------
IF X>0
Let
Therefore, g is strictly increasing.
We can conclude that there exists a>0 such that
So
The function f is decreasing if x<a and then increasing if x>a. Thus f(a) is a minimum of f(x), for x>0.
Since , we have
, which is since
Sorry for the confusion For some reason, I took at least twice the wrong function, and hence the *woops* mistakes