How do you solve sin(e^x)=0 for x?
The question is "find the smallest number x such that sin(e^x)=0".... but I can't even get AN answer lol!
Much appreciated!
well, sine is zero at zero radians, and at PI radians
e^x=0
because e^x cannot be zero, then
e^x=PI take the ln of each side
x=ln(PI)
Well, solving equations can be a sine of trouble sometimes! But fear not, I'm here to help you clown around with the solution!
To find the smallest number x for which sin(e^x) = 0, we need to find the value of x that makes the sine function equal to zero.
Here's the fun part: the sine function equals zero when its input is any integer multiple of pi. So, we need to find all the values of x that satisfy this condition in our equation sin(e^x) = 0.
Now, e^x represents the exponential function, which is always positive (unless you give it a case of the negatives, but let's not go there). The only way for sin(e^x) to equal zero is if e^x is an integer multiple of pi.
So, x must be an integer multiple of ln(pi). Therefore, the answer is x = n * ln(pi), where n is an integer.
But remember, the "smallest" value depends on whether you're hanging out in the land of negative or positive x. So, if you want the smallest positive solution, x = 0. And if you're interested in negative values, well, the party starts with x = -ln(pi).
Now, go forth and may all your trigonometric equations be as smooth as a circus act!
To solve the equation sin(e^x) = 0 for x, you need to find the values of x that satisfy this equation.
Here are the steps to solve it:
1. Start with the equation sin(e^x) = 0.
2. Since sin(x) = 0 when x is any integer multiple of π, you can set the argument inside the sine function equal to an integer multiple of π: e^x = nπ, where n is an integer.
3. Take the natural logarithm of both sides to solve for x: x = ln(nπ).
4. However, keep in mind that for sin(e^x) = 0, you are looking for the smallest value of x. The smallest value of x would occur when n = 0, which makes e^x = 0π = 0.
5. Take the natural logarithm of both sides to get the answer: x = ln(0).
6. However, ln(0) is undefined, which means there is no real number that satisfies sin(e^x) = 0.
Therefore, there is no real value of x that satisfies the equation sin(e^x) = 0.
To solve the equation sin(e^x) = 0 for x, you need to find the values of x that make the sine of e^x equal to zero.
To begin, you can recall that the sine function is zero when the input is a multiple of pi. So, you need to solve the equation e^x = n*pi, where n is an integer.
To isolate x, you can take the natural logarithm of both sides, since the natural logarithm is the inverse function of e^x. This will give you x = ln(n*pi).
However, it's important to note that the number n can be any integer. So, the set of possible solutions is x = ln(n*pi), where n is an integer.
To find the smallest value of x that satisfies the equation, you can substitute different values of n in the equation x = ln(n*pi) and see which one gives the smallest x. Start with n = 1 and continue increasing the value of n until you find the smallest x that satisfies the equation.
For example:
- For n = 1, x = ln(1*pi) = ln(pi)
- For n = 2, x = ln(2*pi)
- For n = 3, x = ln(3*pi)
- ...
Keep substituting larger values of n until you find the smallest x that satisfies the equation sin(e^x) = 0.