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Evaluate the expression: lne^e
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ln e 0 e 1
first use the property of the log that \(\Large \ln a^b = b \ln a \)
then you can simply plug in ln e =1
so the answer is 1?
or 0
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no, did you first use the porperty i gave ?
\(\ln a^b = b \ln a \\ \ln e^e = ... ?\)
honestly dont get that formula haha
wait lne^e = elne
yes, correct! and with ln e = 1 what do u get ?
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e * 1?
which = e
correct! :)
Thanks! :D
welcome ^_^
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