Log help please (1 Viewer)

[ghabi]

Is there any way to solve this type of question without guess and check??

Log(x) = x - 1


Or this one... i must be missing an important rule. any help would be much appreciated

1 + 2x(logx) = 0

x = ???

 

[pLuvia]

You would probably have to do it graphically, or by estimation using Newton-Raphson method or the bisection method

 

[Forbidden.]

You would probably have to do it graphically, or by estimation using Newton-Raphson method or the bisection method

Raphson ? I thought it was just approximation by Newton's method

 

[pLuvia]

Raphson ? I thought it was just approximation by Newton's method

Same thing, Newton method=Newton-Raphson method

 

[kony]

for the first one:

lnx = x-1,
x = e^(x-1) -->taking base e of both sides
ex = e^x -->multiply by e on both sides

by inspection, x = 1.

to prove that there are no further solutions:

d/dx{lnx} = 1/x. therefore, for x>1, the gradient of lnx is less than 1.
d/dx{x-1} = 1, which is always 1.

therefore, the gradient of y=x-1 is always more than lnx for x>1.

hence, there are no further intersections.

similarly, the gradient of lnx is always >1 for x<1, and hence also will never have an intersection.