Given relations are
#log_2 (f(x)) = log_2 (2+2/3+2/9... oo) . log_3 (1+(f(x))/(f(1/x)))# and #f(10) = 1001#
when
#log_2 (f(x)) = log_2 [2(1+1/3+1/3^2... oo) ] log_3 (1+(f(x))/(f(1/x)))#
#=>log_2 (f(x)) = log_2 [2xx(1/(1-1/3)) ] log_3 (1+(f(x))/(f(1/x)))#
#=>log_2 (f(x)) = log_2 [2xx(1/(2/3)) ] log_3 (1+(f(x))/(f(1/x)))#
#=>log_2 (f(x)) = log_2 3 log_3 (1+(f(x))/(f(1/x)))#
#=>log_2 (f(x)) = log_2 (1+(f(x))/(f(1/x)))#
#=>f(x) = 1+(f(x))/(f(1/x))#
#=>f(10) = 1+(f(10))/(f(1/10))#
Now inserting #f(10) = 1001#
#=>1001 = 1+1001/(f(1/10))#
#=>f(1/10)=1001/1000=1+1/1000=1+(1/10)^3#
Inserting #1/10=y# we get
#f(y)=1+y^3#
When #y=20#
#f(20)=1+20^3=8001#