find the solution of this ln equation:
ax=lnx
i tried:
$\displaystyle
e^{(ax)}=x
$
$\displaystyle
e^{(ax)}-x=0
$
what to do next??
i thought of building a taylor series around 0 for ln
but ln(0) is undefined
??
find the solution of this ln equation:
ax=lnx
i tried:
$\displaystyle
e^{(ax)}=x
$
$\displaystyle
e^{(ax)}-x=0
$
what to do next??
i thought of building a taylor series around 0 for ln
but ln(0) is undefined
??
Why not define $\displaystyle f(x) = \ln x -ax$ which iscontinuous on $\displaystyle (0,\infty)$ with $\displaystyle \lim_{x \to 0^+} f(x) = - \infty$ and $\displaystyle \lim_{x \to \infty} = - \infty$. Then consider $\displaystyle f' = \frac{1}{x} - a$ to find the max. If the max is < 0, the no solutions, = 0, then 1 solution and > 0, then 2 solutions.
if i will find the extreme point
f'(x)=0
$\displaystyle
f'(x)=\frac{1}{x}-a
$
$\displaystyle
\frac{1}{x}=a
$
then
$\displaystyle
x=\frac{1}{a}
$
i need to put points $\displaystyle x1>\frac{1}{a}$ and $\displaystyle x2<\frac{1}{a}$
and if f'(x2)>0 and f'(x1)<0
then its a maximum point
i cant see in your explanation any thing that could tell how to get the number of solution regarding the finding of a maximum point
??
(this function is not a parabola we build a delta >0 in order to deside that there are two solutions)
You're right. The max is located at $\displaystyle x = \frac{1}{a}$ and so the actual maximum value is
$\displaystyle f_{\text{max}}= f\left( \frac{1}{a}\right) = - \ln a -1$
Now clearly if $\displaystyle a < \frac{1}{e}$ then $\displaystyle f_{\text{max}}> 0 $ and since f is continuous on $\displaystyle (0,\infty)$ with f approaching negative infinity as $\displaystyle x \to 0^{+},\; x \to \infty$ then there are x values say $\displaystyle x_0,\; x_1$ where $\displaystyle x_1 < \frac{1}{e} < x_2$ where $\displaystyle f(x_1) < 0,\;f(x_2) < 0$ and so by the intermediate value theorem there exist $\displaystyle c_1, \;c_2,\;c_1 \ne c_2$ where $\displaystyle f(c_1) = 0,\;f(c_2) = 0$ giving two solutions to your problem. Similary for $\displaystyle a = \frac{1}{e}$ and $\displaystyle a > \frac{1}{e}$
the intermediate value theorem says that if
there are two points for which f(x1)<0 and f(x2)>0
then there is a x point between then which is a solution
f(x)=0
i found only one such point in the maximum point
there is no another case where f(x1)<0 and f(x2)>0
from where you got the second solution fact??
our function is
f(x)=ln(x)-ax
our extreme point is x=1/a
f''(x)=-1/x^2 if f''(x1)>0 X1 is minimum if f''(x2)<0 X2 is maximum
f(1/a)=ln(1/a)-1
we cant know if its minum or maximum because it depends on the value of
parameter a.
we cant assume that our derivative will get a possitive or negative value
when we input to the second derivative a parameter
so i am confused regarding your solution