# Thread: Assignment Logarithmic issue (solved past logs but can't solve last bit)

1. ## Assignment Logarithmic issue (solved past logs but can't solve last bit)

Hello MHF! First time poster here, the following is my problem:

$2 log(x) + log(x-1) = 1$

I solved the above by grouping the logs together, resulting in the below:

$log[x^2(x-1)] = 1$

$log[x^3-x^2] = 1$

By removing logs I ended up with:

$x^3 - x^2 = 10$

I simply have no idea how to solve this (it's probably simpler than I might imagine) but the x cubed is confusing me and I cannot find anything I can pull out in common that lets me solve with with an answer of 10 for both brackets or whatever.

Any help is appreciated on the matter.

2. ## Re: Assignment Logarithmic issue (solved past logs but can't solve last bit)

Since this equation doesn't have any rational solutions there are a number of different ways of finding the solution for x. Here are some suggestions:
Interval Bisection
Cubic Formula (not so common but takes a form similar to the Quadratic Formula)
Newton-Raphson method

In any case this equation cannot be solved analytically but we can see that a solution exists between 2 < x < 3. Hope this has given you some ideas

3. ## Re: Assignment Logarithmic issue (solved past logs but can't solve last bit)

Originally Posted by OllieC
Since this equation doesn't have any rational solutions there are a number of different ways of finding the solution for x. Here are some suggestions:
Interval Bisection
Cubic Formula (not so common but takes a form similar to the Quadratic Formula)
Newton-Raphson method

In any case this equation cannot be solved analytically but we can see that a solution exists between 2 < x < 3. Hope this has given you some ideas

I do recall something along the lines of a solution involving a range, but I can't really figure out the proper method to do so, could you please explain in a little more detail?

Edit: Does it by any chance involve testing several numbers replacing x and figure out which ones provides a result closest to 0?

4. ## Re: Assignment Logarithmic issue (solved past logs but can't solve last bit)

The Bisection Method is the easiest to apply. Decide a region which your solution lies in, halve it, then see which half the solution lies in and repeat the process until you have your desired accuracy.

5. ## Re: Assignment Logarithmic issue (solved past logs but can't solve last bit)

Thanks kindly!