We can evaluate that answer without calculators, here is the trick, W(ln4) = W(ln2²) = W(2ln2) = W(e^ln2 * ln2) = ln2. So, lnx = ln2, from here, x = e^ln2 = 2.
@@tambuwalmathsclass Yes of course, you have to use a super advanced calculator to compute that. But do you know that l can approximate it without these machines ?
Can you give us the method you usually use instead of Lambert w function?
Graphical solution or guess and check
We can evaluate that answer without calculators, here is the trick,
W(ln4) = W(ln2²) = W(2ln2) = W(e^ln2 * ln2) = ln2.
So, lnx = ln2, from here, x = e^ln2 = 2.
@@ChidexOnlineMathClass01 What if the number is 3.7
@@tambuwalmathsclass
3.7 is different and it's a decimal number.
@@ChidexOnlineMathClass01 There you have to use a machine that understands Lambert W function
@@tambuwalmathsclass
Yes of course, you have to use a super advanced calculator to compute that. But do you know that l can approximate it without these machines ?
I have a calculator with the function
@@ChidexOnlineMathClass01 Where can I get it?
What's the name of the calculator
Ah 🎉yuo a a Nigerian
Yes