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प्रश्न
Express in terms of log 2 and log 3 :
`"log"26/51 - "log"91/119`
उत्तर
`"log"26/51 - "log"91/119`
= `"log"((26/51)/(91/119)) ....[ log_am - log_an = log_a(m/n)]`
= `"log"26/51 xx 119/91`
= `"log" [ 2 xx 13 ]/[ 3 xx 17 ] xx [ 7 xx 17 ] /[ 7 xx 13 ]`
= `"log" 2/3`
= log 2 - log 3 ......[ loga`m/n` = logam - logan ]
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