# Homework Help: Solve Logx^2

1. Feb 5, 2010

### storoi1990

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

2. Relevant equations

3. The attempt at a solution

Q: Logx = 1 + 6/logx

my attempt so far: Log x^2 = 7

where do i go from here?

2. Feb 5, 2010

Re: Logx^2

How did you obtain this result? Try to multiply the whole equation with logx. You'll arrive at a quadratic equation. Try to use a substitution then.

3. Feb 5, 2010

### HallsofIvy

Re: Logx^2

Is this $log x= 1+ \frac{6}{log x}$ or $log x= \frac{1+ 6}{log x}$?

I suspect it is the former because otherwise it would just be written as 7/ log x. But then multiplying log x= 1+ (6/log x) by log x you get (log x)^2= log x+ 6. Let y= log x and that becomes y^2= y+ 6. Solve that quadratic equation for y, then solve log x= y for x.

4. Feb 5, 2010

### storoi1990

Re: Logx^2

yeah it's the last one.

so then i end up with:

y^2-y-6 = 0

x1 = 3 , and x2 = -2

is that the answer?

5. Feb 5, 2010

### Staff: Mentor

Re: Logx^2

You mean the first one, which is logx = 1 + (6/logx).