How do you solve #\ln x - \ln ( x - 2) = \ln 8#?

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Answer 1 · 2017-02-02T08:17:31

Use the property #ln(a)-ln(b) = ln(a/b)# on the left side.
Make the logs disappear by using the exponential function on both sides.
Solve for x.

Given: #ln(x) - ln(x-2) = ln(8)#

Use the property #ln(a)-ln(b) = ln(a/b)# on the left side.

#ln(x/(x-2)) = ln(8)#

Use the exponential function on both sides:

#e^(ln(x/(x-2))) = e^(ln(8))#

Because the exponential function and the natural logarithm are inverses, they cancel:

#cancel(e)^(cancel(ln)(x/(x-2))) = cancele^(cancel(ln)(8))#

This is the equation with the cancelled functions removed.

#x/(x-2) = 8#

Multiply both sides by #x-2#

#x= 8(x-2)#

#x = 8x-16#

#-7x=-16#

#x = 16/7#

Check #ln(16/7)-ln(16/7-2) = ln(8)# with a calculator.

#2.07944#(and more digits)# = 2.07944#(and more digits)

Therefore, it checks.