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Find the solution ln(x-1)>1 for x>1
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x-1>e x>e+1
any questions?
Nope thank you very much!!!!
first change to experiential from \[e^{\ln(x-1)}>e ^{1}\] x-1>e x>e+1, since e+1 is always greater then 1, there is only one solution of x>e+1
Thanks a lot,it is part of a function for me to solve and needen some help,!
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let me now if you need anymore help gl
Just a note: if you have an inequality x > y, then\[e^x > e^y\]is also true because the exponential is a monotonically increasing function. This is not true for any function (for example,\[\sin(x) > \sin(y)\]may not be true).
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