Question

lne^5

Answer 1

"e" is a constant defined by:
\[\lim_{x \rightarrow \infty} (1+\frac{ 1 }{ x })^{x}\]
but, you'll better know it as:
\[e=2.718281\]
The expression "ln" is a logarithm that has a base of "e" so:
\[lnx=\log_{e} x\]
Now, having that out of the way, let's look at the problem in question:
\[lne^5\]
If you remember from your algebra classes, there is a property of the logarithms that involves an exponent in the logarithmic number:
\[\log_{b} a^n = n \log_{b} a\]
So let's apply it with the problem:
\[lne^5=5lne\]
Since I said that "ln" is a logarithm with the base "e", then by logarithmic definition applied, I know that lne=1, so I'll sustitude it on the problem:
\[5lne=5(1)\]
and now, it is solved, so I can now be sure that lne^5 = 5 :
\[lne^5=5\]