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anonymous
 5 years ago
In(x+1)In(x)=1 i know the answer is 1over e1 but how?
anonymous
 5 years ago
In(x+1)In(x)=1 i know the answer is 1over e1 but how?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I rearranges it: ln(x+1)=1+ln(x) Substitute 1=ln(e) and use the properties of logarithms to get ln(x+1)=ln(ex) So x+1=ex Solve for x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I recommend this way , too \[e^{\ln(x+1)\ln(x)}=e^1\] by taking both sides to the exponent

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so by property of exponents, \[{e^{\ln(x+1)} \over e^{\ln(x)} } =e\] and by definition of logarithm, \[{x+1 \over x} = e\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so by cross multiplication x+1 = ex xex = 1 x(1e) = 1 x =\[1 \over 1e\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you multiply 1/1 to make the top = 1, \[1 \over e1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let me know if there is anything that you didn't get

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hey Yuki if you could please elaborate a little on this part  " \[e^{\ln(x+1)}\div e^{\ln(x)}=e\] and by definition of logarithm, x+1/x = e" How do you end up on the second equation from the first?
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