Hello there. To solve this simultaneous equation, we'll have to remember some properties about logarithms.
Given the equations:
[tex]\begin{gathered} \log _2(6x)=1+2\log _2(y) \\ 1+\log _6(x)+\log _6(15y-25) \end{gathered}[/tex]First, we'll transform those ones into logarithms applying the following rule in reverse:
[tex]\log _a(a)=1[/tex]Thus, we get:
[tex]\log _2(6x)=\log _2(2)+2\log _2(y)[/tex]For the first equation; Now, apply the power and product rules in reverse:
[tex]\begin{gathered} \log _c(a^b)=b\cdot\log _c(a) \\ \log _c(ab)=\log _c(a)+\log _c(b) \end{gathered}[/tex]In order to obtain:
[tex]\log _2(6x)=\log _2(2)+\log _2(y^2)=\log _2(2y^2)[/tex]Since both logarithms have the same base, the expressions inside them must be equal as well, therefore:
[tex]\begin{gathered} 6x=2y^2 \\ \\ y^2=3x \end{gathered}[/tex]Now, going for the second equation, we'll apply the same three rules as before:
[tex]\begin{gathered} 1+\log _6(x)=\log _6(15y-25) \\ \log _6(6)+\log _6(x)=\log _6(15y-25) \\ \log _6(6x)=\log _6(15y-25) \\ 6x=15y-25 \end{gathered}[/tex]The thing is that we can plug in 6x as y², found on the first equation, to get a quadratic equation for y:
[tex]2y^2=15y-25[/tex]Subtract 15y-25 on both sides of the equation
[tex]2y^2-15y+25=0[/tex]Using the quadratic formula, we take the roots of the equation:
[tex]y=\frac{15\pm\sqrt[]{225-4\cdot2\cdot25}}{2\cdot2}=\frac{15\pm\sqrt[]{25}}{4}=\frac{15\pm5}{4}[/tex]So we have two possible roots:
y = (15 - 5)/4 = 10/4 = 5/2 or y = (15 + 5)/4 = 20/4 = 5.
Plugging in this value in the first equation, we solve for x:
[tex]\begin{gathered} 6x=2\cdot\mleft(\frac{5}{2}\mright)^2 \\ 6x=2\cdot\frac{25}{4} \\ x=\frac{25}{12} \end{gathered}[/tex]Or
[tex]\begin{gathered} 6x=2\cdot5^2 \\ 6x=2\cdot25=50 \\ x=\frac{25}{3} \end{gathered}[/tex]The pair of solutions are:
[tex]\begin{gathered} (x,y)=\mleft(\frac{25}{3},5\mright) \\ (x,y)=\left(\frac{25}{12},\frac{5}{2}\right) \end{gathered}[/tex]You can plug this into any of the equations and see if they truly satisfy them as an exercise.
