Solve the equation #ln(x)+ln(x-3)+ln(2x-6)=0#?

Question

2772 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-30T00:14:32

# x ~~ 3.3844 #

We have:

# ln(x)+ln(x-3)+ln(2x-6)=0 #

If we look at the graph of:

# y= ln(x)+ln(x-3)+ln(2x-6) #
graph{ln(x)+ln(x-3)+ln(2x-6) [-10, 10, -5, 5]}

Then we appear to have a solution, #alpha ~~ 3.5#

First note that for each individual logarithm to exist we require #x# to simultaneously satisfy the following inequalities:

# ln(x) in RR => x gt 0 #
# ln(x-2) in RR => x-3 gt 0 => x gt 3#
# ln(2x-5) in RR => 2x-6 gt 0 => x gt 3#

Thus #x gt 3#

We can find this solution algebraically:

# ln(x)+ln(x-3)+ln(2x-6)=0 #

# :. ln{x(x-3)(2x-6)}=0 #
# :. x(x-3)(2x-6) = e^0 #
# :. x(x-3)(2x-6) = 1 #

And we can multiply out the expression to get:

# x(2x^2-6x-6x+18) = 1 #
# :. 2x^3-12x^2+18x = 1 #
# :. 2x^3-12x^2+18x - 1 = 0#

As this cubic does not factorise we solve it numerically:

graph{2x^3-12x^2+18x - 1 [-10, 10, -5, 5]}

And we get three real solutions:

# x ~~ 0.0578, 2.5579, 3.3844 #

From earlier we established that #x gt 3#, thus we can eliminate two of these solutions leaving:

# x ~~ 3.3844 #