Advertisements
Advertisements
प्रश्न
Compute log9 27 – log27 9
उत्तर
log927 – log279 = log9 33 – log27 32
= 3 log9 3 – 2 log27 3 ......(1)
By change of base rule [logb a = `1/log_"a""b"`]
(1) ⇒ log927 – log279 = `3/log_3 9 - 2/log_3 27`
= `3/log_3 3^2 - 2/log_3 3^3`
= `3/(2log_3 3) - 2/(3log_3 3)`
= `3/2 - 2/3` ....[∵ log3 3 = 1]
= `(9 - 4)/6`
= `5/6`
APPEARS IN
संबंधित प्रश्न
Let b > 0 and b ≠ 1. Express y = bx in logarithmic form. Also state the domain and range of the logarithmic function
Solve log8x + log4x + log2x = 11
Prove `log "a"^2/"bc" + log "b"^2/"ca" + log "c"^2/"ab"` = 0
Prove that `log 2 + 16log 16/15 + 12log 25/24 + 7log 81/80` = 1
Prove that loga2 a + logb2 b + logc2 c = `1/8`
Prove log a + log a2 + log a3 + · · · + log an = `("n"("n" + 1))/2 log "a"`
If `log x/(y - z) = logy/(z - x) = logz/(x - y)`, then prove that xyz = 1
Solve log5 – x (x2 – 6x + 65) = 2
Choose the correct alternative:
The value of `log_(sqrt(5))` 512 is
Choose the correct alternative:
The value of `log_3 1/81` is
Choose the correct alternative:
If `log_(sqrt(x)` 0.25 = 4, then the value of x is
Choose the correct alternative:
The value of logab logbc logca is
Choose the correct alternative:
If 3 is the logarithm of 343, then the base is
Choose the correct alternative:
The value of log3 11 . log11 13 . log13 15 . log15 27 . log27 81 is
