hardest math problem ever?

Question

Find x > 3 such that

ln(x) < x^(0.1)"

Answer 1

gjhnfjnjn

Answer 2

ln(x) < x^(0.1) and x>3

So, since e^a function is increasing , this means that if a>y implies that e^a > e^y.

Set a = x^(0.1) and y = ln(x) , By applying the inequality above together with the "exponential function" we get by the reasoning above,

e^ [ x^(0.1) ] > e^ [lnx] and note that e^[lnx] = x :) So,
e^ [ x^(0.1) ] > x

Just "try for some values of x"... >3...Even '3' works!
e^ [ x^(0.1) ] = e^ [ 3^(0.1) ] = 3.05.. > 3 = x

Thus the above is true for ALL x>3.... since the "increasing nature" of the function e^ [ x^(0.1) ] - x which I will leave it as an exercise for you... trick is to differentiate twice and show that this second derivative is >0 for all x>3.

Hope this helped.

:)

Answer 3

Can´t be solved algebraically
Take
y=lnx-x^0.1
At x=0 lim y= -infinity and at x ==> +infinity the same
y´= 1/x-0.1 x^-0.9 =0 so 1-0.1x^0.1=0
x^0.1=10 taking log (base10)
0.1log x= 1 so log x = 10 and x=10^10
at this point
y(10^10) = 10ln10 -10 >0 so there are two zeros a<10^10 and b>10^10
solution 0b
Use any method to approximate the roots

Answer 4

Are you sure that's the problem?

Because there are many, many values of x that satisfy that equation.

e.g. e^1000

ln x = 1000
x^0.1 = e^100, obviously bigger than 1000

Answer 5

3.059726680 is the intersections, so
3 < x < 3.059726680
The only way i can think to do it is to graph the two lines and find the intersection.

They also intersect at 3.430631121x10^15, so x could also be x > 3.430631121x10^15