Question 30534: a)given that 3 + 2log2 x= log2 y show that y=8x2
b)Find the roots α and β where α<β of the equation
3+ 2log2 x = log2 (14x-3)
c)Show that log2 α = -2
d)calculate log2 β, giving your answer to 3sf.
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! a)given that 3 + 2log2 x= log2 y show that y=8x2
3+2LOG(X)/LOG(2) = LOG(Y)/LOG(2)...SINCE LOG(X) TO BASE A = LOG(X)/LOG(A)..TO ANY COMMON BASE.
3LOG(2)+2LOG(X)=LOG(Y)
LOG(2^3)+LOG(X^2)=LOG(Y)...SINCE LOG(X^N)=N*LOG(X)
LOG(8)+LOG(X^2)=LOG(Y)
LOG(8X^2)=LOG(Y)...SINCE LOG(X*Y) = LOG(X)+LOG(Y)
TAKING ANTILOGS
Y=8X^2
b)Find the roots α and β where α<β of the equation
3+ 2log2 x = log2 (14x-3)
USING RESULT OF ABOVE PROBLEM ,OR PROCEEDING SAME WAY AS ABOVE,WE GET..
14X-3=8X^2
8X^2-14X+3=0
8X^2-12X-2X+3=0
4X(2X-3)-1(2X-3)=0
(2X-3)(4X-1)=0
ALPHA=1/4....AND.....BETA=3/2....
c)Show that log2 α = -2
LOG(1/4)TO BASE 2 =LOG(1/2^2) TO BASE 2..
LOG(2^-2)TO BASE 2 = -2
d)calculate log2 β, giving your answer to 3sf.
LOG(3/2) TO BASE 2.=LOG(1.5)/LOG(2)
=0.5849 =0.585
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