Solve the simultaneous equations ${log}_{a} (x + y) = 0$ $log_a(x+y) = 0$ and $2 {log}_{a} x = log (4 y + 1)$ $2log_ax = log(4y +1)$ for $x$ $x$ and $y$ $y$ ?

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Answer 1 · 2017-08-15T14:55:50

Solution is $(- 2 + \sqrt{7}, 3 - \sqrt{7})$ $(-2+sqrt7,3-sqrt7)$

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

Further $2 {log}_{a} x = {log}_{a} (4 y + 1) \Rightarrow {log}_{a} x^{2} = {log}_{a} (4 y + 1)$ $2log_ax=log_a(4y+1)=>log_ax^2=log_a(4y+1)$ (B)

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

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

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

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

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

and using quadratic formula

$y = \frac{6 \pm \sqrt{36 - 8}}{2} = 3 \pm \sqrt{7}$ $y=(6+-sqrt(36-8))/2=3+-sqrt7$

If $y = 3 + \sqrt{7}$ $y=3+sqrt7$ , $x = - 2 - \sqrt{7}$ $x=-2-sqrt7$ , but we cannot have $x < 0$ $x<0$

and if $y = 3 - \sqrt{7}$ $y=3-sqrt7$ , $x = - 2 + \sqrt{7}$ $x=-2+sqrt7$

Hence solution is $(- 2 + \sqrt{7}, 3 - \sqrt{7})$ $(-2+sqrt7,3-sqrt7)$

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