# Math Help - sum of roots...

1. ## sum of roots...

find Sum of Roots of the equation $4x^{\lfloor log_{10}2x-3 \rfloor} = 1$

Where $\lfloor x \rfloor$ = greatest integer function.

like $\lfloor 2.3 \rfloor = 2$

2. $4x^{\lfloor \log_{10}2x-3 \rfloor} = 1$
$\log_{10}(4x^{\lfloor \log_{10}2x-3 \rfloor}) = \log_{10}(1)$
$\log_{10}4 + \log_{10}(x^{\lfloor \log_{10}2x-3 \rfloor})) = 0$
$\log_{10}4 + \lfloor \log_{10}2x-3 \rfloor \log_{10}(x) = 0$
$\log_{10}4 + \lfloor \log_{10}2x-3 \rfloor (\log_{10}2x - \log_{10}2) = 0$

Let $y = \log_{10}2x$
$\log_{10}4 + \lfloor y-3 \rfloor (y - \log_{10}2) = 0$

First solve without greatest integer (to get close to the solution)
$\log_{10}4 + (y-3) (y - \log_{10}2) = 0$
$y^2 - (3 + \log_{10}2)y + 5 \log_{10}2 = 0$
Solutions are around 2.75 and 0.55

Now with this we guess values for $a = \lfloor y\rfloor$ (0,1,2,3)
$\log_{10}4 + (a - 3) (y - \log_{10}2) = 0$
$y = -\frac{\log_{10}4}{a - 3} + \log_{10}2$
We see that $a = 0$ is the only one with a consistent solution.
$y = \frac{\log_{10}4}{3} + \log_{10}2 = \log_{10}2^{\frac{5}{3}}$
$y = \log_{10}2x$
$x = \frac{10^y}{2} = 2^{\frac{2}{3}}$

And this is the only solution for $x$