How do you solve #9= - 1+ \sqrt { 10\ln - 1}#?

1 Answer
Alan N.
Jun 22, 2017

#x=e^10+1 approx 22027.4657948#
[If the equation was supposed to be #9=-1+sqrt(10ln(x-1)#]

Explanation:

Clearly the question as posed is incomplete as there is no variable.
I will assume the equation was supposed to be #9=-1+sqrt(10ln(x-1)# and solve for #x#.

#9=-1+sqrt(10ln(x-1)#

#sqrt(10ln(x-1)) = 10#

Square both sides

#10ln(x-1) =100#

#ln(x-1) =10#

#(x-1) = e^10#

#x= e^10+1 approx 22027.4657948#

NB: Another reasonable assumption would be: #9=-1+sqrt(10lnx-1)#

This would be solved in the same way: #x=e^10.1#

There are, of course, countless others!

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