Does ln((e^u)+(e^v))=u+v or uv?
Thanks for clarifying!
neither, you are confused in the basic law:
ln(xy) = lnx + lny ≠ ln(x+y)
testing your result,
let v = 2, u = 3
then ln(e^3 + e^2)
= ln(27.47745..)
= appr 3.31
≠ 3+2 nor 2*3
What would ln(e^u+e^v) equal then?
To determine whether ln((e^u)+(e^v)) equals u+v or uv, we can use the properties of logarithms.
First, we have ln((e^u)+(e^v)). Let's simplify this expression step by step.
The expression inside the natural logarithm can be rewritten as a single exponential expression using the property of logarithms:
ln(e^u) represents the exponent u, and ln(e^v) represents the exponent v. So, we have:
ln((e^u)+(e^v)) = ln(e^u) + ln(e^v)
As per the logarithmic property, ln(a) + ln(b) = ln(a * b), we can simplify further:
ln((e^u)+(e^v)) = ln(e^u * e^v)
Since e^u * e^v is a single term, we can simplify it using the property e^(a+b) = e^a * e^b:
ln((e^u)+(e^v)) = ln(e^(u+v))
Now, we can see that the logarithm and exponential functions cancel each other out:
ln((e^u)+(e^v)) = u + v
Therefore, ln((e^u)+(e^v)) equals u + v, not uv.
Remember to apply the properties of logarithms and exponentials when dealing with these types of expressions.