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masterchiefo
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Homework Statement
1. I have to prove that x+ln(1+e^(-x)) is equal to ln(1+e^(x))
Homework Equations
The Attempt at a Solution
1. x+ln(1+e^(-x)) =ln(1+e^(x))
ln(e^(-x)+1)+x
ln(1/(e^x)+1)+x
Im blocked here.. :(
thanks
that a and b is equal, same numberPeroK said:Hint:
If ##e^a = e^b## what can you say about ##a## and ##b##?
masterchiefo said:that a and b is equal, same number
This is not the best way to start out, since this is what you're supposed to prove.masterchiefo said:Homework Statement
1. I have to prove that x+ln(1+e^(-x)) is equal to ln(1+e^(x))
Homework Equations
The Attempt at a Solution
1. x+ln(1+e^(-x)) =ln(1+e^(x))
masterchiefo said:ln(e^(-x)+1)+x
ln(1/(e^x)+1)+x
Im blocked here.. :(
thanks
posted an attempt in my last postehild said:You need to know that ##e^{-x}=\frac{1}{e^x}## and ##\ln(\frac{a}{b})=\ln(a)-\ln(b)##.
write 1+e-x in fraction form.
It is right.masterchiefo said:so 1 + e^(-x) becomes 1 + 1/e^x which we can make one fraction with a common denominator of e^x. so it'll be (e^x/e^x) + (1/e^x)
or (e^x + 1)/(e^x). we can use props of lns to change that into ln(e^x + 1) - ln(e^x)
so it'd be x + ln(e^x + 1) - ln(e^x) = ln(1 + e^x)
for ln(e^x) the ln and e cancel each other, so only the x stays
so it'd be x + ln(e^x + 1) - x = ln(e^x + 1)
and the x and -x cancel each other. so you have ln(e^x+1) = ln (e^x+1)
is this right ?
awesome man thank you very much.ehild said:It is right.
masterchiefo said:Sorry guys I still don't understand how to do this, been reading online tutorials on how to solve ln equation with no luck.
I guess you saw no value in paying attention to the equation I wrote in post # 7. If you substitute this into your original equation, you get:masterchiefo said:awesome man thank you very much.
yeah sorry I did not really understand what your post meant :/Chestermiller said:I guess you saw no value in paying attention to the equation I wrote in post # 7. If you substitute this into your original equation, you get:
$$x+\ln (1+e^{-x})=\ln e^x+\ln (1+e^{-x})=\ln [e^x(1+e^{-x})]$$
What does that give you?
Chet
The definition of lny is the power to which you have to raise e to get y. Now let's substitute ex for y in the previous statement:masterchiefo said:yeah sorry I did not really understand what your post meant :/
I am not very good with ln /e/log solver.
A logarithm is a mathematical function that represents the exponent needed to produce a certain number. In this problem, the logarithm is used to solve for the value of x that satisfies the given equation.
One way to prove this is to use the properties of logarithms and algebraic manipulation to transform one side of the equation into the other. Another way is to use a graphing calculator or software to graph both sides of the equation and see that they intersect at the same point, indicating that they are equal.
No, there are certain restrictions on the values of x that can be used. In this case, x must be a real number and cannot be equal to zero or any other value that would make the natural logarithms undefined.
The purpose of the natural logarithm is to solve for the value of x that satisfies the equation. It is used to cancel out the exponential terms and isolate the variable on one side of the equation.
Yes, there are other methods that can be used to solve this equation, such as substitution, elimination, and graphing. However, using logarithms is the most common and efficient way to solve problems involving exponential and logarithmic functions.