Logarithms

Since, $7-\log _3|x| \geq 0 \Rightarrow \log _3|x| \leq 7$

And $4-\log _3|x| \geq 0 \Rightarrow \log _3|x| \leq 4$

$ \Rightarrow|x| \leq 3^4=81 $

Now, $7-\log _3|x|=\left(4-\log _3|x|\right)^2$

$ \begin{aligned} & \Rightarrow 7-\log _3|x|=16-8 \log _3|x|+\left(\log _3|x|\right)^2=0 \\ & \Rightarrow\left(\log _3|x|\right)^2-7 \log _3|x|+9=0 \end{aligned} $

Let $t=\log _3|x|$, then

$ \begin{aligned} & t^2-7 t+9=0 \\ & \Rightarrow t=\frac{7 \pm \sqrt{49-36}}{2}=\frac{7 \pm \sqrt{13}}{2} \end{aligned} $

Since, $t=\log _3|x|$ must be a real number and we know that $0 \leq \log _3|x| \leq 4$

So, only roots satisfy the condition is

$ 0 \leq \frac{7-\sqrt{13}}{2} \leq 4 $

and this root provides two solutions for $x$.

$[\because|x|$ means both $x$ and $-x$ satisfying the equation $]$

Thus, the number of real solutions is 2 .

${\log _{225}}70 = {{{{\log }_e}70} \over {{{\log }_e}225}} = {{{{\log }_e}7 + {{\log }_e}5 + {{\log }_e}2} \over {{{\log }_e}25 + {{\log }_e}9}}$

$ = {{1 + {{{{\log }_e}5} \over {{{\log }_e}7}} + {{{{\log }_e}2} \over {{{\log }_e}7}}} \over {2{{{{\log }_e}5} \over {{{\log }_e}7}} + 2{{{{\log }_e}3} \over {{{\log }_e}7}}}}$

$ = {{1 + {{\log }_7}5 + {{\log }_3}2 \times {{\log }_5}3 \times {{\log }_7}5} \over {2[{{\log }_7}5 + {{\log }_5}3 \times {{\log }_7}5]}}$

$ = {{1 + a + abc} \over {2[a + ab]}} = {{1 + a + abc} \over {2a(1 + b)}}$

We have,

${\log _5}{{(a + b)} \over 3} = {{{{\log }_5}a + {{\log }_5}b} \over 2}$

$ \Rightarrow 2{\log _5}\left( {{{a + b} \over 3}} \right) = {\log _5}ab$

$ \Rightarrow {\log _5}{\left( {{{a + b} \over 3}} \right)^2} = {\log _5}ab$

$ \Rightarrow {{{{(a + b)}^2}} \over 9} = ab$

$ \Rightarrow {(a + b)^2} = 9ab$

$ \Rightarrow {a^2} + {b^2} = 7ab$

$ \Rightarrow {a \over b} + {b \over a} = 7$

$ \Rightarrow {\left( {{a \over b} + {b \over a}} \right)^2} = 49$

$ \Rightarrow {{{a^2}} \over {{b^2}}} + {{{b^2}} \over {{a^2}}} + 2 = 49$

$ \Rightarrow {{{a^4} + {b^4}} \over {{a^2}{b^2}}} = 49 - 2 = 47$