Does E X have a solution?

Originally Answered: Is there any solution for e^x=x? Slope of e^x is 1 at x = 0 and it keeps increasing after that. Thus y = x cannot cut y = e^x. Thus no real solution.

How do you solve EXX 5?

1. Transform the equation ex+x=5 into ex=5−x.
2. Write each side of the new equation as function : y1=ex and y2=5−x.
3. Step.5 The x-value is the solution of our equation : x=1.307 (approximately)

Is X X no solution?

The solution x = 0 means that the value 0 satisfies the equation, so there is a solution. “No solution” means that there is no value, not even 0, which would satisfy the equation. This is because there is truly no solution—there are no values for x that will make the equation 12 + 2x – 8 = 7x + 5 – 5x true.

How do you solve by elimination?

To Solve a System of Equations by Elimination

1. Write both equations in standard form.
2. Make the coefficients of one variable opposites.
3. Add the equations resulting from Step 2 to eliminate one variable.
4. Solve for the remaining variable.
5. Substitute the solution from Step 4 into one of the original equations.

Which equation has no real solution?

quadratic equation
A quadratic equation has no solution when the discriminant is negative. From an algebra standpoint, this means b2 < 4ac. Visually, this means the graph of the quadratic (a parabola) will never touch the x axis. Of course, a quadratic that has no real solution will still have complex solutions.

Is there any solution for e^x=x?

No there is no real solution for the above equation. See this graph for more clarity. Both the curves are not cutting at any point. Originally Answered: Is there any solution for e^x=x? Slope of e^x is 1 at x = 0 and it keeps increasing after that. Thus y = x cannot cut y = e^x.

Does the equation $x>1$ have a unique solution?

The function $\\log x$is increasing and continuous for $x\\ge 1$. The values of these functions at $1$are $1/e$and $0$. The values of these functions at $e$are $e^{-e}<\\log e=1$. So yes, the equation has unique solution $x>1$.

Can Y = X cut Y = E^X?

Slope of e^x is 1 at x = 0 and it keeps increasing after that. Thus y = x cannot cut y = e^x. Thus no real solution. What does Google know about me?

What are the solutions to E^z = 0?

Hrm? e^z = 0 has no solutions, real or complex. The clause in Picard’s theorem “…attains every value infinitely often with at most one exception” is usually demonstrated using e^z, which has an essentially singularity at infinity but you will never find it to be equal to zero anywhere near there (or anywhere else, of course).