Tasty Logarithm #3

let $\log _{ 225 }{ x }$ = $p$ and $\log _{ 64 }{ y }$ = $q$

$p$ + $q$ =4

$\frac { 1 }{ p }$ - $\frac { 1 }{ q }$ =1

solving both the equations we obtain

$p$ = $3\pm \sqrt { 5 }$ , $q$ = $1\pm \sqrt { 5 }$

${ x }_{ 1 }$ = ${ 225 }^{ 3+\sqrt { 5 } }$ , ${ x }_{ 2 }$ = ${ 225 }^{ 3-\sqrt { 5 } }$ , ${ y }_{ 1 }$ = ${ 64 }^{ 1+\sqrt { 5 } }$ , ${ y }_{ 2 }$ = ${ 64 }^{ 1-\sqrt { 5 } }$ .

${ x }_{ 1 }{ x }_{ 2 }{ y}_{ 1 }{ y }_{ 2 }$ = ${ 225 }^{ 3+\sqrt { 5 } }.{ 225 }^{ 3-\sqrt { 5 } }.{ 64 }^{ 1+\sqrt { 5 } }.{ 64 }^{ 1-\sqrt { 5 } }$ = ${ 225 }^{ 6 }.{ 64 }^{ 2 }$ = ${15 }^{ 12 }.{ 2 }^{ 12 }$ = ${ 30 }^{ 12 }$

$\log _{ 30 }{ { \left( { { x }_{ 1 }{ x }_{ 2 }{ y }_{ 1 }{ y }_{ 2 } } \right) } }$ = $\log _{ 30 }{ { \left( { { 30 }^{ 12 } } \right) } }$ = $\huge\boxed { 12 }$

Let $log_{225} x = a$ and $log_{64} y= b$

$a+b = 4$

$\frac{1}{a} - \frac{1}{b}=1$

Solving we have the two quadratic equations, we have:

$a^2 - 6a +4 = 0$ and $b^2 - 2b - 4 = 0$

So we have $a_1 + a_2 = 6$ while $b_1 + b_2 = 2$

This means:

$log_{225} x_1 + log_{225} x_2 = 6$ and $log_{64} y_1 + log_{64} y_2 = 2$

$log_{225} (x_1 x_2) = 6$ and $log_{64} (y_1 y_2) = 2$

Hence $x_1 x_2 = 225^6$ and $y_1 y_2 =64^2$

$log_{30} (x_1 x_2 y_1 y_2) = log_{30} (225^6 64^2) = log_{30} ((15^2)^6 (2^6)^2)$

$= log_{30} (15^{12} 2^{12}) = log_{30} (15 \times 2)^{12} = log_{30} (30^{12}) = \boxed{12}$ .