Advertisements
Advertisements
Question
Prove that `("log"_"p" x)/("log"_"pq" x)` = 1 + logp q
Solution
L.H.S.
= `("log"_"p" x)/("log"_"pq" x)`
= `((("log" x)/("log""p")))/((("log"x)/("log""pq"))`
= `("log"x)/("log""p") xx ("log""pq")/("log"x)`
= `("log""pq")/("log""p")`
= `("log""p" + "log""q")/("log""p")`
= `1 + ("log""q")/("log""p")`
= 1 + logp q
= R.H.S.
Hence proved.
APPEARS IN
RELATED QUESTIONS
If `3/2 log a + 2/3` log b - 1 = 0, find the value of a9.b4 .
If log`( a - b )/2 = 1/2( log a + log b )`, Show that : a2 + b2 = 6ab.
Evaluate: logb a × logc b × loga c.
Solve the following:
log(x2 + 36) - 2log x = 1
Solve the following:
log ( x + 1) + log ( x - 1) = log 11 + 2 log 3
Solve for x: `("log"125)/("log"5)` = logx
If 2 log x + 1 = log 360, find: log(2 x -2)
Express the following in a form free from logarithm:
m log x - n log y = 2 log 5
Express the following in a form free from logarithm:
5 log m - 1 = 3 log n
Prove that: `(1)/("log"_8 36) + (1)/("log"_9 36) + (1)/("log"_18 36)` = 2