SOLUTION: Inx + In(x-2) = 1 The answer should be: 1 + Square root of 1+e =2.928

Question 414844: Inx + In(x-2) = 1

The answer should be: 1 + Square root of 1+e =2.928
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
ln(x) + ln(x-2) = 1
Solving equations where the 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)

With the "non-log" term of 1, the first form will be easier to achieve than the "all-log" second form. All we have to do is find a way to combine the two logarithms into one.

They are not like terms so we cannot just add them. (Like logarithmic terms have the same bases and same arguments. You logarithms have the same base, e, but the arguments, x and x-2, are different.)

Fortunately there is a property of logarithms, , which allows us to combine logarithms

of the same base
with coefficients of 1
and a "+" sign between them.

Your logarithms meet all these requirements. So we can use the property to combine your logarithms:
ln(x*(x-2)) = 1
which simplifies to:

We noe have the first form. The next step with the first form is to rewrite the equation in exponential form. In general is equivalent to . Using this pattern on your equation we get:

which simplifies to:

Now we solve this equation. This is a quadratic equation so we want one side to be zero. Subtracting e from each side we get:

Now we factor or use the Quadratic Formula. With the "e" this does not factor so we must use the Quadratic Formula:

which simplifies as follows:

In long form this is:
or

When solving this kind of equation you must check your answer(s). You must ensure that each "solution" makes all arguments of all logarithms positive. Any "solution" that makes any argument of any logarithm zero or negative must be rejected. These rejected solutions can occur even if no mistakes were made! So even expert mathematicians must check their answers on these problems.

Use the original equation to check:
ln(x) + ln(x-2) = 1
Checking :

is a positive number which is greater than 1. So is a positive number greater than two. Therefore both arguments will end up positive. So there is no reason to reject this solution. This is the required part of the check. Completing the check will tell us if we made a mistake. You are welcome to finish the check.

Checking :

is a positive number which is greater than 1. So will be a negative number. Since is negative and since it is the argument of the first logarithm, we must reject this solution. (It doesn't matter what happens with the second logarithm.)

So the only solution to your equation is

This is an exact expression for the solution. Use your calculator if you want a decimal approximation.
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