Consider the logarithm function $log=exp^{-1}$ on $\mathbb{R}$. Prove:
d) $log : {\mathbb{R}}_{+} \rightarrow \mathbb{R} $ is strictly monotonically increasing, and thus injective.

Question

Consider the logarithm function $log=exp^{-1}$ on $\mathbb{R}$. Prove:
d) $log : {\mathbb{R}}_{+} \rightarrow \mathbb{R} $ is strictly monotonically increasing, and thus injective.

experimentx · Answer

$$ f(x) = \ln x\
f'(x) = {1 \over x} \
f'(R_+) = +ve \text{ hence it is increasing} $$

experimentx · Answer

for any two +ve number 'a' and 'b' such that a = b
exp^-1(a) = exp^-1(b) => log(a) = log(b)
and for any two real numbers log(a) = log(b) => e^log(a) = e^log(b) => a = b

since it is strictly increasing function (sinve the slope > 0), we can say that it is injective function

anonymous · Answer

thanks experimentX $$ln x$$ is the reverse of $$e^{x}$$ right ?

experimentx · Answer

inverse

anonymous · Answer

aah ok sorry my english not so good :) thank you

experimentx · Answer

yw