How do you solve #ln sqrt(x-8)= 5#?

1 Answer
Jun 21, 2016

Answer:

We go step by step un-doing the things that are being done to the variable, #x#, making sure that we do the same thing to both sides arriving at the answer:

#x=e^10+8#

Explanation:

We go step by step un-doing the things that are being done to the variable, #x#, making sure that we do the same thing to both sides. The first thing we encounter is the natural logarithm, which we can un-do using it's inverse, #e^x#. Starting with the left hand side:

#e^ln(sqrt(x-8))=sqrt(x-8)#

then the right hand side is:

#e^5#

Which makes our equation:

#sqrt(x-8)=e^5#

Now we un-do the square-root by squaring both sides. Starting with the left hand side:

#(sqrt(x-8))^2=x-8#

and the right hand side:

#(e^5)^2=e^(5*2)=e^10#

which makes our equation:

#x-8=e^10#

Now we can add #8# to both sides giving:

#x=e^10+8#