Logarithms question. Very hard.?
回答 (8)
2009-05-22 9:16 am
✔ 最佳答案
2 log3 Y - log3 (Y + 4) = 2log3(Y^2) - log3(Y + 4) = 2
log3(Y^2/(Y+4)) = 2
9 = Y^2/(Y+4)
Y^2 = 9(Y+4)
Y^2 - 9Y - 36 = 0
(Y - 12)(Y + 3) = 0
Y - 12 = 0
Y + 3 = 0
Y = 12, -3
Discard Y = -3 because the log of a negative number is undefined, so the answer is Y = 12
2009-05-22 5:21 pm
2log_3(y) - log_3(y + 4) = 2
log_3(y^2) - log_3(y + 4) = 2
log_3[(y^2)/(y + 4)] = 2
(y^2)/(y + 4) = 3^2
y^2 = 9(y + 4)
y^2 = 9y + 36
y^2 - 9y - 36 = 0
y^2 + 3y - 12y - 36 = 0
(y^2 + 3y) - (12y + 36) = 0
y(y + 3) - 12(y + 3) = 0
(y + 3)(y - 12) = 0
y + 3 = 0
y = -3 (unacceptable since it's a negative no.)
y - 12 = 0
y = 12
â´ y = 12
log_3(y^2) - log_3(y + 4) = 2
log_3[(y^2)/(y + 4)] = 2
(y^2)/(y + 4) = 3^2
y^2 = 9(y + 4)
y^2 = 9y + 36
y^2 - 9y - 36 = 0
y^2 + 3y - 12y - 36 = 0
(y^2 + 3y) - (12y + 36) = 0
y(y + 3) - 12(y + 3) = 0
(y + 3)(y - 12) = 0
y + 3 = 0
y = -3 (unacceptable since it's a negative no.)
y - 12 = 0
y = 12
â´ y = 12
2009-05-22 4:54 pm
Let log be log to base 3
2 log Y - log ( Y + 4 ) = 2
log Y^2 + log (Y + 4 )^(-1) = 2
log [ Y^2 / (Y + 4 ) ] = 2
Y^2 / (Y + 4 ) = 9
Y^2 = 9 Y + 36
Y^2 - 9 Y - 36 = 0
( Y - 12) ( Y + 3 ) = 0
Y = 12 is acceptable
2 log Y - log ( Y + 4 ) = 2
log Y^2 + log (Y + 4 )^(-1) = 2
log [ Y^2 / (Y + 4 ) ] = 2
Y^2 / (Y + 4 ) = 9
Y^2 = 9 Y + 36
Y^2 - 9 Y - 36 = 0
( Y - 12) ( Y + 3 ) = 0
Y = 12 is acceptable
2009-05-22 4:21 pm
log (base 3) y^2/y+4=2 =====>
y^2/y+4=9
y^2=9y+36
y^2-9y-36=0
y=[9 + & - (81+144)^1/2]/2=12,-1/2 not accepable.
y^2/y+4=9
y^2=9y+36
y^2-9y-36=0
y=[9 + & - (81+144)^1/2]/2=12,-1/2 not accepable.
2009-05-22 4:18 pm
2 log3 Y - log3 (Y + 4) = 2
=> log3 {y^2 / (y+4)} = 2 => y^2 = 9(y+4) => y^2 – 9y – 36 = 0
=> (y-12)(y+3) = 0 => y = 12, (y can not be negative)
=> log3 {y^2 / (y+4)} = 2 => y^2 = 9(y+4) => y^2 – 9y – 36 = 0
=> (y-12)(y+3) = 0 => y = 12, (y can not be negative)
2009-05-22 4:18 pm
3^(LHS) = 9
y^2 / (y+4) = 9
y^2 - 9y - 36 = 0
(y-12)(y+3) = 0
y = 12
y^2 / (y+4) = 9
y^2 - 9y - 36 = 0
(y-12)(y+3) = 0
y = 12
2009-05-22 4:16 pm
2log3 Y - log3 (Y + 4) = 2
log3 (Y²) - log3 (Y + 4) = 2
log3 (Y²)/(Y + 4) = 2
(Y²)/(Y + 4) = 3²
(Y²)/(Y + 4) = 9
Y² = 9(Y + 4)
Y² = 9Y + 36
Y² - 9y - 36 = 0
(Y - 12)(Y + 3) = 0
Y = 12 {only take positive value, ignore the negative value because it is in logarithm form}
log3 (Y²) - log3 (Y + 4) = 2
log3 (Y²)/(Y + 4) = 2
(Y²)/(Y + 4) = 3²
(Y²)/(Y + 4) = 9
Y² = 9(Y + 4)
Y² = 9Y + 36
Y² - 9y - 36 = 0
(Y - 12)(Y + 3) = 0
Y = 12 {only take positive value, ignore the negative value because it is in logarithm form}
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