Advertisements
Advertisements
प्रश्न
Prove that : `2"log" 15/18 - "log"25/162 + "log"4/9 = log 2 `
उत्तर
We need to prove that
`2"log"15/18 - "log"25/162 + "log"4/9 = log 2`
LHS = `2"log"15/18 - "log"25/162 + "log"4/9`
= `"log"(15/18)^2 - "log"(25/162) + "log"(4/9)` ....[ nlogam = logamn ]
= `"log"[(15/18) xx (15/18)] - "log"25/162 + "log"4/9`
= `"log"(15/18) xx (15/18) xx (4/9) - "log"(25/162) .....[ log_am + log_an = log_a(mn)]`
= `"log"((15/18) xx (15/18) xx 4/9)/(25/162) .....[ log_am - log_an = log_a(m/n)]`
= `"log" (15/18) xx (15/18) xx 4/9 xx 162/25`
= `"log" 72/36`
= log 2
= R.H.S.
APPEARS IN
संबंधित प्रश्न
Express in terms of log 2 and log 3 : log 4.5
Find x, if : x - log 48 + 3 log 2 = `1/3`log 125 - log 3.
Solve for x : log10 (x - 10) = 1
Solve for x : log (x2 - 21) = 2.
If log 2 = 0.3010 and log 3 = 0.4771; find the value of : log 25
If log 2 = 0.3010 and log 3 = 0.4771; find the value of:
`2/3` log 8
If log102 = a and log103 = b ; express each of the following in terms of 'a' and 'b': log 12
If log102 = a and log103 = b; express each of the following in terms of 'a' and 'b' : log 60
If log102 = a and log103 = b; express each of the following in terms of 'a' and 'b' : log `3 1/8`
State, true or false :
`log x/log y` = log x - log y