Question

Prove that: `(1)/("log"_8 36) + (1)/("log"_9 36) + (1)/("log"_18 36)` = 2

Answer 1

L.H.S.
= `(1)/("log"_8 36) + (1)/("log"_9 36) + (1)/("log"_18 36)` 

= log₃₆ 8 + log₃₆ 9 + log₃₆ 18

= `("log"8)/("log"36) + ("log"9)/("log"36) + ("log"18)/("log"36)`

= `(1)/("log"36)("log" 8 + "log"9 + "log"18)`

= `(1)/("log"36)("log"2^3 + "log"3^2 + "log"(2 xx 3^2))`

= `(1)/("log"(2^2 xx 3^2))("log"2^3 + "log"3^2 + "log"2 + "log"3^2)`

= `(1)/("log"(2^2 xx 3^2))(3"log"2 + 2"log"3 + "log"2 + "log"3)`

= `(1)/(2"log"2 + 2"log"3)(4"log"2 + 4"log"3)`

= `(4)/(2("log"2 + "log"3))("log"2 + "log"3)`
= 2
= R.H.S.
Hence proved.