# Equation with logarithmic and polynomial terms

1. Feb 6, 2013

### NanakiXIII

This is not actually a homework question, but it seemed appropriate to put it here. In an old exam from 1921 I found the following problem. I never learned how to solve this type of thing and I haven't been able to figure it out, so: how does one solve this?

1. The problem statement, all variables and given/known data

Solve for $x$:

$$\frac{(x-1)^2}{(x-1)-^6\log (x-1)} = 3 \times 6^{3\times^6\log 2 + 2\times^6\log 3}$$

2. Relevant equations

3. The attempt at a solution

I went ahead and simplified this to

$$y^2 -72y + 72 ^6\log y=0$$

where $y=x-1$, but, as I said, I never learned how to solve this type of equation involving both polynomial terms and logarithms and I don't know how to proceed.

2. Feb 6, 2013

### Staff: Mentor

You have several instances of expressions such as 6log (something). I suspect that these really mean log6(something). IOW, the log expressions are log-base 6. Please clarify.

3. Feb 6, 2013

### NanakiXIII

Yes, exactly. That's how we wrote log-base-6 in school.

4. Feb 6, 2013

### HallsofIvy

Staff Emeritus
On the right, use the laws of logarithms: $$a log(x)= log(x^a)$$, log(x)+ log(y)= log(xy), and $$b^{log_b(x)}= x$$ to get $$3(6^{3log_6(2)+ 2log_6(3)})= 3(6^{log_6((2^3)(3^2)})= 3(8)(9)= 216$$

By the way, it is NOT a good idea to use "$\times$" to indicate multiplication when you have $x$ as the unknown. Just use parentheses.

5. Feb 7, 2013

### NanakiXIII

Yes, it would appear I miscalculated, the 72's in my equation should be 216's. I'll edit my post. That said, though, I still don't know how to solve the equation.

Edit: Actually, it appears I can't edit the original post, so here's an erratum:

My simplification (last equation) should be

$$y^2 - 216 y - 216 \log_6 y = 0$$

where $y = x-1$.

Last edited: Feb 7, 2013
6. Feb 7, 2013

### HallsofIvy

Staff Emeritus
Since that equation involves both powers of y and logarithm of y, the solution cannot be written in terms of "elementary" functions. It should be possible, by taking the exponential of both sides, to get it in the form $ve^v= \text{constant}$ and then solve it in terms of the Lambert W function.

7. Feb 10, 2013

### NanakiXIII

Well, that explains why I never learned to solve this. However, since this is an old high school exam, I would be surprised if there was not a simpler solution. It is probably possible to solve it using elementary functions and a logarithm table.