Question #2641d

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Answer 1 · 2018-02-06T14:36:12

#1/x#

Remember that #a^b=e^(b*lna)#.

Here, #a=x#, and #b=-1#.

The above equation becomes #e^(-1*lnx)#.

Using the above property of powers, this simplifies to:

#x^-1#

#=1/x#

Answer 2 · 2018-02-06T14:38:27

The answer is #=1/x#

Let #y=e^(-lnx)#

Taking the logarithms on both sides,

#lny=ln(e^(-lnx))#

#=-lnx#

#=ln1-lnx#

#=ln(1/x)#

As,

#lny=ln(1/x)#

#y=1/x#