Advertisements
Advertisements
प्रश्न
Prove that `("log"_"p" x)/("log"_"pq" x)` = 1 + logp q
Advertisements
उत्तर
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
संबंधित प्रश्न
If log√27x = 2 `(2)/(3)` , find x.
Solve for x, `log_x^(15√5) = 2 - log_x^(3√5)`.
Solve for x: `("log"27)/("log"243)` = x
Solve for x: `("log"125)/("log"5)` = logx
If log 3 m = x and log 3 n = y, write down
32x-3 in terms of m
If log 3 m = x and log 3 n = y, write down
`3^(1-2y+3x)` in terms of m an n
If a = `"log" 3/5, "b" = "log" 5/4 and "c" = 2 "log" sqrt(3/4`, prove that 5a+b-c = 1
Prove that log (1 + 2 + 3) = log 1 + log 2 + log 3. Is it true for any three numbers x, y, z?
If a b + b log a - 1 = 0, then prove that ba.ab = 10
If log (a + 1) = log (4a - 3) - log 3; find a.
