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प्रश्न
Prove that logbm a = `1/"m" log_"b""a"`
उत्तर
logbm a = `1/"m" log_"b""a"`
L.H.S. = logbm a
= `log"a"/log "b"^"m" ...[log_y x = logx/logy]`
= `log"a"/("m"log"b")`
= `1/"m"log_"b""a"`
= R.H.S.
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