1
Honey Arora
·2009-07-12 21:05:26
The given equation can be written as
log 3/log x =2x-3.
log 3=(2x-3)log x log 3=log(x)2x-3
i.e. 3=x2x-3
I couldn't find any value of x satisfying this equation
1
yes no
·2009-07-12 22:04:26
PROCEDURE to draw the GRAPH of x^(2x-3)
possible if x>0 (ok??)
at x =1 it is one
at x = 1.5 it is one hence between x =1 and x=1.5 it takes a turn..
also at x = 0 it is infinite
and for higher values of x it approaches to infinity..right?
after x=1.5 it will always increase??clear??(no need to do dy/dx)..just try and observe its behaviour by putting values
similarly for x<1
(two solutions)
24
eureka123
·2009-07-12 22:44:02
@ yes
PLZZZZZZZZZZ DONT POST COMPUTER GENERATED GRAPHS[16][16]
WE ARE TRYING HERE TO IMPROVE OUR SOLVING SKILLS ONLY.....SOO PLZZZDONT REPEAT IT[1][1]
1
Honey Arora
·2009-07-13 01:59:47
i hv drawn this from #2 and #4
1
yes no
·2009-07-13 04:12:20
yes honey is right
gud..that is correct graph of x^(2x-3)
11
Mani Pal Singh
·2009-07-13 07:52:13
Sorry Honey i didn't saw ur earlier posts
[1]
But 4m next time do write continued!