- #1
Luhter
- 8
- 0
How do you solve this?
[itex] e^x+2x-5=0 [/itex]
And i mean the algorithm
[itex] e^x+2x-5=0 [/itex]
And i mean the algorithm
arildno said:Apart from the minor advantage of already-tabulated approximate values of the Lambert function, I don't see any pedagogical advantage in that approach.
I love complex analysis as well, not the least its beauty, and how it <i>clarifies</i> and organizes seemingly disparate issues.jackmell said:I just like complex analysis even though I'm not very good at it. It has made concepts in math more clear for me and I think it would clear up things for other students as well.
The first step in solving this equation is to bring the constant term to the right side by adding 5 to both sides, resulting in e^x+2x=5.
To isolate the exponential term, we can take the natural logarithm (ln) of both sides, resulting in ln(e^x+2x)=ln(5). This simplifies to x+ln(e^2)=ln(5).
After isolating e^x, we can simplify the natural logarithm by using the property ln(e^y)=y, which means ln(e^2)=2. This leaves us with x+2=ln(5).
To solve for x, we can subtract 2 from both sides, resulting in x=ln(5)-2. This gives us the exact value of x that satisfies the equation e^x+2x-5=0.
Yes, this equation can be solved using a calculator. However, it is important to know the steps and concepts involved in solving it algebraically, as it is a fundamental part of understanding exponential functions and logarithms.