# Solve the simultaneous equations log_a(x+y) = 0 and 2log_ax = log(4y +1) for x and y ?

Aug 15, 2017

Solution is $\left(- 2 + \sqrt{7} , 3 - \sqrt{7}\right)$

#### Explanation:

As ${\log}_{a} \left(x + y\right) = 0$, $x + y = 1$ as log of $1$ is always zero. (A)

Further $2 {\log}_{a} x = {\log}_{a} \left(4 y + 1\right) \implies {\log}_{a} {x}^{2} = {\log}_{a} \left(4 y + 1\right)$ (B)

i.e. ${x}^{2} = 4 y + 1$

as from A, $x = 1 - y$, putting this in B, we get

${\left(1 - y\right)}^{2} = 4 y + 1$

or $1 - 2 y + {y}^{2} = 4 y + 1$

or ${y}^{2} - 6 y + 2 = 0$

$y = \frac{6 \pm \sqrt{36 - 8}}{2} = 3 \pm \sqrt{7}$
If $y = 3 + \sqrt{7}$, $x = - 2 - \sqrt{7}$, but we cannot have $x < 0$
and if $y = 3 - \sqrt{7}$, $x = - 2 + \sqrt{7}$
Hence solution is $\left(- 2 + \sqrt{7} , 3 - \sqrt{7}\right)$