1.Ta có:log2580=12log580=12(1+4log52)b=log√2740=2log2740=2(log278+log275)=2(log32+13log35)a=log950=12log350=12(log32+2log25)⇒{log32=3b−2a5log35=12a−3b10⇒log52=log32log35=6b−4a12a−3b⇒log2580=12+12b−8a12a−3b
2.Ta có:
log2580=12log580=12(1+4log52)b=log√2740=2log2740=2(log278+log275)=2(log32+13log35)a=log950=12log350=12(log32+2log25)⇒{log32=3b−2a5log35=12a−3b10⇒log52=log32log35=6b−4a12a−3b⇒log2580=12+12b−8a12a−3b