How Do You Solve Logarithmic Equations Involving Quadratics?

[storoi1990]

Q: Logx = 1 + 6/logx

my attempt so far: Log x^2 = 7

where do i go from here?

 

[radou]

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.

 

[HallsofIvy]

Q: Logx = 1 + 6/logx

my attempt so far: Log x^2 = 7

where do i go from here?

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.

 

[storoi1990]

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?

 

[Mark44]

yeah it's the last one.

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

so then i end up with:

y^2-y-6 = 0
[\quote]This factors into (y - 3)(y + 2) = 0, so y = 3 or y = -2.

Now, since y = logx, you have logx = 3 or log x = -2.
What are the solutions for x? They are NOT 3 and -2, as you show below.

x1 = 3 , and x2 = -2

is that the answer?