Question #04717

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Answer 1 · 2017-03-23T13:16:01

Given: #ln((x-5)/(x-1))=ln(6)#

Subtract #ln(6)# from both sides:

#ln((x-5)/(x-1))-ln(6)= 0#

Use the property of logarithms where subtracting is equivalent to division within the argument:

#ln((x-5)/(6(x-1)))= 0#

Make both sides the exponent of e:

#e^(ln((x-5)/(6(x-1))))= e^0#

The logarithm and the exponential on the left side cancel and the right side becomes 1:

#(x-5)/(6(x-1))= 1#

Now, it is easy to solve for x:

#x-5=6x-6#

#5x=1#

#x = 1/5#

Check:

#ln((1/5-5)/(1/5-1))=ln(6)#

#ln((1-25)/(1-5))=ln(6)#

#ln((-24)/(-4))=ln(6)#

#ln(6) = ln(6)#

This checks.

#x = 1/5#

Answer 2 · 2017-03-24T00:20:43

#x=1/5#

The contents of the two logarithms, which are the only things present on either side of the equation, are equal:

#(x-5)/(x-1)=6#

#=>x-5=6(x-1)#

#=>x-5=6x-6#

#=>1=5x#

#=>x=1/5#