1. ## Logarithmic Equations

Hi, so this gets me nowhere, especially after I get rid of the logarithm and I get 57/2. Maybe I'm missing something and you can point it out, this is what I have:

log5x + (log5125/log5x) = 7/2
log5x(125 - x) = 7/2
log5(125x - x2) = 7/2
125x - x2 = 57/2
125x - x2 = 279.5
x2 - 125x - 279.5 = 0

Trial and error is no good, and quadratic seems to be the only option.

Thanks

2. ## Re: Logarithmic Equations

Let $u=\log_5 x$

Then $u+\dfrac{3}{u}=\dfrac{7}{2}$

$2u^2-7u+6=0$

$u=\dfrac{3}{2}$ or $u=2$

$x=\sqrt{125}$ or $x=25$.

3. ## Re: Logarithmic Equations

Originally Posted by kaperdomo
Hi, so this gets me nowhere, especially after I get rid of the logarithm and I get 57/2. Maybe I'm missing something and you can point it out, this is what I have:
log5x + (log5125/log5x) = 7/2
$y+\dfrac{3}{y}=\dfrac{7}{2}$ where $y=\log_5(x)$ therefore $x=y^5$
Can you solve for $y~?$
$y+\dfrac{3}{y}=\dfrac{7}{2}$ where $y=\log_5(x)$ therefore $x=y^5$ . . . No.
Can you solve for $y~?$
Therefore $\ \ x = 5^y.$