SOLUTION: The question states: Solve each logarithmic equation
1.) log(3x-1)=log(x-3)+log6
How do i solve this?
Algebra.Com
Question 542445: The question states: Solve each logarithmic equation
1.) log(3x-1)=log(x-3)+log6
How do i solve this?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the basic logarithm concepts used to solve this problem are:
log(a) + log(b) = log(a*b)
log(a) - lob(b) = log(a/b)
log(a) = b if and only if 10^b = a
note that log(a) means log(a) to the base of 10.
if the base is 10, then it doesn't have to be shown.
if the base is anything other than 10, then it has to be shown.
for example:
log(5,a) means log of a to the base of 5.
you will use the 3 concepts shown above to solve this problem.
your equation is:
log(3x-1)=log(x-3)+log(6)
you use the first concept to transform the right side of this equation to get:
log(3x-1) = log((x-3)*6) which becomes:
log(3x-1) = log(6x-18)
you then subtract the expression on the right side of the equation from both sides of the equation to get:
log(3x-1) - log(6x-18) = 0
you then use the second concept to transform the left side of this equation to get:
log((3x-1)/(6x-18)) = 0
you then apply the third concept to transform this equation to get:
log((3x-1)/(6x-18)) = 0 if and only if:
10^0 = (3x-1)/(6x-18)
since 10^0 equals 1, then this equation becomes:
(3x-1)/(6x-18) = 1
you then multiply both sides of this equation by (6x-18) to get:
3x-1 = 6x-18
you then add 18 to both sides of this equation and subtract 3x from both sides of this equation to get:
-1+18 = 6x-3x
you then combine like terms to get:
17 = 3x
you then divide both sides of this equation by 3 to get:
x = 17/3
that's your answer.
you then substitute for x in your original equation and solve to confirm that this answer is good.
i did so using my calculator and confirmed that it's good.
my hand written calculations are shown below:

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