Tuesday, January 6, 2009
Challenging question!!?
Solve
1)log (5x + 8) + log (2x +1) - 2log (x +1) = 1
Show working for this question. Thanx!
[(5x+8)(2x+1)]^1/(x+1)^2 = 10
(5x+8)(2x+1) = 10 (x+1)^2
10x^2 + 5x +16x + 8 = 10x^2 +20x + 10
21x-20x = 2
x=2
Therefore, (5x+8)(2x+1)/(x+1)^2 =10
10x^2+21x+8 = 10x^2 +20x +10
x = 2
log [(5x+8)(2x+1)] - log[(x+1)^2] = 1
log (10x^2 + 21x + 8) - log(x^2 + 2x + 1) = 1
log[ (10x^2 + 21x + 8) /(x^2 + 2x + 1)] = 1 = log 10
(10x^2 + 21x + 8) /(x^2 + 2x + 1) = 10
10x^2 + 21x + 8 = 10x^2 + 20x + 10
x = 2
Not very challenging
=> log {(5x +8) * (2x +1)} - log (x + 1)^2 = log 10
=> log {(5x +8) * (2x +1} / (x + 1)^2 = log 10
=> {(5x +8) * (2x +1} / (x + 1)^2 = 10
=> 10x^2 + 21x + 8 = 10*(x^2 + 2x + 1)
=> 10x^2 + 21x + 8 = 10x^2 + 20x + 10
=> x = 2
2log(x+1)=log[(x+1)^2]
Then
log (5x + 8) + log (2x +1) - 2log (x +1)=log (5x+8)(2x+1)-log[{x+1}^2]
= log [ (5x+8)(2x+1)/(x+1)^2]
If log Y=1 then Y=10 so
(5x+8)(2x+1)/(x+1)^2=10
(5x+8)(2x+1)=10(x+1)^2
10x^2+21x+8=10x^2+20x+10
x=2
Check: left hand side =log18 +log 5 -2log3=log(18*5)-log3^2
=log90-log9=log(90/9)=log10=1
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Posted by jane |