Question

Solve for x
lnx +ln(x-1) = 1

Answer 1

Start by using both sides as exponents for e.

Answer 2

ln a + ln b = ln ab

Answer 3

e ^(a + b) - (e^a)(e^b). And e^(ln x) = x

Answer 4

ln(x)(x-1) =1

Answer 5

Is what I have so far good? Where do I go from here?

Answer 6

e^[lnx +ln(x-1)] = e^1 = e. [e^(ln x)][e^(ln [x-1])] = x(x-1) = e = x^2 - x. Now, just solve quadratic.

Answer 7

I tried solving it and I got x = 2.22 a,d x = -1.22 and I was told the negative could not be used so I used 2.22 and when I typed it in to my online math thing it said it was wrong

Answer 8

and*

Answer 9

and x > 1 because the original equation has an ln (x-1), so x-1 has to be positive.

Answer 10

never mind I u=guess I wasn't putting in enough digits, it is correct, thanks!

Answer 11

you're welcome

Answer 12

Hey do this:

\[lnx +ln(x-1) = 1\]

Re-write the equation as follows:

\[ln(x(x-1)) =1\]
\[ln(x^2 - x) = 1\]

Take the inverse ln of both sides to get

\[x^2 - x = e\]

Subtract e from both sides to get

\[x^2 - x - e = 0\]

Now use the quadratic formula

where a = 1, b = -1, c = -e to solve for x