Advertisements
Advertisements
Question
Prove that `("log"_"p" x)/("log"_"pq" x)` = 1 + logp q
Advertisements
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 log2(x + y) = log3(x - y) = `log 25/log 0.2`, find the values of x and y.
Given log10x = 2a and log10y = `b/2`. Write 10a in terms of x.
Evaluate : log38 ÷ log916
Evaluate : `( log _5^8 )/(( log_25 16 ) xx ( log_100 10))`
Solve the following:
log (x + 1) + log (x - 1) = log 48
Solve for x: `("log"27)/("log"243)` = x
If log x = a and log y = b, write down
10a-1 in terms of x
If 2 log x + 1 = log 360, find: x
If x + log 4 + 2 log 5 + 3 log 3 + 2 log 2 = log 108, find the value of x.
Prove that (log a)2 - (log b)2 = `"log"("a"/"b")."log"("ab")`
