Advertisements
Advertisements
Question
Solve the following:
`log_2x + log_4x + log_16x = (21)/(4)`
Solution
`log_2x + log_4x + log_16x = (21)/(4)`
∴ `(1)/("log"_x2) + (1)/("log"_x2^2) + (1)/("log"_x2^4) = (21)/(4)`
∴ `(1)/("log"_x2) + (1)/("2log"_x2) + (1)/("4log"_x2) = (21)/(4)`
∴ `(1)/("log"_x2) (1 + 1/2 + 1/4) = (21)/(4)`
∴ `(1)/("log"_x2)(7/4) = (21)/(4)`
∴ `"log"_x2 = (7)/(4) . (4)/(21)`
∴ `"log"_x2 = (1)/(3)`
∴ `x^(1/3)` = 2
∴ x = 23
= 8.
APPEARS IN
RELATED QUESTIONS
Evaluate :`1/( log_a bc + 1) + 1/(log_b ca + 1) + 1/ ( log_c ab + 1 )`
If x = log 0.6; y = log 1.25 and z = log 3 - 2 log 2, find the values of :
(i) x+y- z
(ii) 5x + y - z
Solve the following:
log 8 (x2 - 1) - log 8 (3x + 9) = 0
Solve the following:
log (x + 1) + log (x - 1) = log 48
Solve for x: `("log"27)/("log"243)` = x
State, true of false:
If `("log"49)/("log"7)` = log y, then y = 100.
If `"log" x^2 - "log"sqrt(y)` = 1, express y in terms of x. Hence find y when x = 2.
If 2 log x + 1 = log 360, find: x
Express the following in a form free from logarithm:
2 log x + 3 log y = log a
Prove that (log a)2 - (log b)2 = `"log"("a"/"b")."log"("ab")`
