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How do you solve #ln(y-1) - ln(2) = x + ln x#?

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1 Answer

Dec 12, 2015

Use property of logs ...

Explanation:

#ln[(y-1)/(2xxx)]=x#

Exponentiate ...

#e^{ln[(y-1)/(2xxx)]}=e^x#

Simplify ...

#(y-1)/(2x)=e^x#

#y=(2x)e^x+1#

This function approaches #y=1# as #xrarr-oo#. It never crosses the x-axis.

Hope that helps

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