Recall that, #a^x=y iff log_ay=x#.
Accordingly, #972^x=8 rArr x=log_972 8#.
By the Change of Base Rule (CBR), then, #x=log_b 8/log_b 972#,
where #b# is a new base, with #b gt 0, b!=1#.
Similarly, #y=log_b 16/log_b 243#.
#:. 3/x-4/y=3*log_b 972/log_b 8-4*log_b 243/log_b 16#,
#=3*log_b 972/log_b 2^3-4*log_b 243/log_b 2^4#,
#=3*log_b 972/(3log_b 2)-4*log_b 243/(4log_b 2)#,
#=log_b 972/log_b 2-log_b 243/log_b 2#,
#=(log_b 972-log_b 243)/log_b 2#,
#=log_b (972/243)/log_b 2#,
#=log_b 4/log_b 2#,
#=log_2 4.................[because," the CBR]"#,
# rArr 3/x-4/y=2#.
