lnx = x^3 + 1

Question

i know x=1 but can you solve this algebraically?

Answer 1

ln(1)=0, which does not equal 0+1
ln(x)-x^3=1 does not have a solution.

Now if you mean lnx=x^3-1, then yes it has two solutions, Since x3 contains the variable to solve for, move it to the left side of the equation by subtracting x3
from both sides.
ln(x)−x3=−1
The roots of this equation could not be found algebraically, so the roots were determined numerically.
x=0.39044,1

Answer 2

numerically on a graphing calculator?

Answer 3

that would be a choice, I would do it on wolfram.

Answer 4

or

Newton's method:
let y = lnx - x^3 + 1
dy/dx = 1/x - 3x^2

newx = x - (lnx - x^3 + 1)/(1/x - 3x^2)
= (1 - 3x^3 - lnx + x^3 - 1)/(1/x - 3x^2)
= (-2x^3 - lnx)/(1/x - 3x^2)

guess: x = .5

newx1 = .3545...
newx2 = .38789...
nrwx3 = .39042..
newx4 = .39044046
newx5 = .39044046 <---- correct to 8 decimals

looked like some fun to do it.