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Question Number 131343 by physicstutes last updated on 03/Feb/21
A mapping is defined as G→S where (G,×) and (S,+), show that the mapping f(x) = ln x is an isomophism.
$$\mathrm{A}\:\mathrm{mapping}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{as}\:{G}\rightarrow{S}\:\mathrm{where}\:\left({G},×\right)\:\mathrm{and}\:\left({S},+\right), \\ $$$$\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{mapping}\:{f}\left({x}\right)\:=\:\mathrm{ln}\:{x}\:\mathrm{is}\:\mathrm{an}\:\mathrm{isomophism}. \\ $$
Answered by mindispower last updated on 04/Feb/21
isomlrphisme is bijection morphisme without knowing G and S we can say just f(1)=0 f(xy)=f(x)+f(y) if x,y>0
$${isomlrphisme}\:{is}\:{bijection}\:\:{morphisme} \\ $$$${without}\:{knowing}\:{G}\:{and}\:{S}\:{we}\:{can}\:{say}\:{just} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${f}\left({xy}\right)={f}\left({x}\right)+{f}\left({y}\right)\:{if}\:{x},{y}>\mathrm{0} \\ $$$$ \\ $$

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