SOLUTION: Could a tutor please help? I need to solve the following out to two decimal places. The problem is: log(2x-1)=log(x+3)+ log3 Thank You so much!!

Question 394319: Could a tutor please help? I need to solve the following out to two decimal places. The problem is:
log(2x-1)=log(x+3)+ log3
Thank You so much!!
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
log(2x-1)=log(x+3)+ log(3)
Solving equations where teh variable is in the argument of a logarithm usually starts with transforming the equation into one of the following forms:
log(expression) = other-expression
or
log(expression) = log(other-expression)

Since your equation has nothing but logarithms the first form will more difficult to achieve. So we will aim for the second form. If we can find a way to combine the two logarithms on the right side into a single logarithm then we would have the second form. Those two logarithms are not like terms so we cannot just add them. But there is a property of logarithms, , which will allow us to combine the two logarithms on the right side:
log(2x-1)=log((x+3)*3)
which simplifies to:
log(2x-1)=log(3x+9)
We now have the equation in the second form. This equation says that the base 10 logarithm of 2x-1 is the same as the base 10 logarithm of 3x+9. The only way these two logarithms can be the same is if the arguments are the same. So:
2x-1 = 3x+9
This is a simple equation to solve. Subtracting 2x from each side we get:
-1 = x + 9
Subtracting 9 from each side and we get:
-10 = x

When solving logarithmic equations like this one, you must check your answers! You must ensure that any "solutions" do not make any argument (or base) of a logarithm negative or zero. This can happen even if not mistakes were made while solving.

If a "solution" does make an argument (or base) of a logarithm negative or zero then that solution must be rejected.

When checking, always use the original equation:
log(2x-1)=log(x+3)+ log(3)
Checking x = -10:
log(2(-10)-1)=log((-10)+3)+ log(3)
which simplifies to:
log(-20-1)=log(-7)+ log(3)
We can now see that an argument (actually two of them) is negative. So we must reject this solution. And since this was the only "solution" we found, your equation has no solution.
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