Advertisements
Advertisements
Question
Solve log8x + log4x + log2x = 11
Solution
log8x = `1/log_x 8 = 1/log_x 2^3 = 1/(3log_x 2)`
log4x = `1/log_x 4 = 1/log_x 2^2 = 1/(2log_x 2)`
log2x = `1/log_x 2`
∴ log8x + log4x + log2x = 11
⇒ `1/(3log_x 2) + 1/(2log_x 2) + 1/log_x 2` = 11
⇒ `1/log_x 2 [1/3 + 1/2 + 1]` = 11
⇒ `1/log_x 2[(2 + 3 + 6)/6]` = 11
⇒ `1/log_x 2 = 11 xx 6/11` = 6
(i.e.,) = 6
x = 26
= 64
APPEARS IN
RELATED QUESTIONS
Compute log9 27 – log27 9
Solve log4 28x = 2log28
If a2 + b2 = 7ab, show that `log ("a" + "b")/3 = 1/2(log"a" + log "b")`
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 `log_2 x − 3 log_(1/2) x` = 6
Solve log5 – x (x2 – 6x + 65) = 2
Choose the correct alternative:
The value of `log_3 1/81` 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
