how to solve log(x-21)=2-log(x)?

2010-03-31 5:51 pm
cant figure it out

回答 (6)

2010-03-31 5:58 pm
✔ 最佳答案
log ( x - 21 ) = 2 - log ( x )

log ( x - 21 ) + log x = 2

log ( x ( x - 21 ) ) = log 10^2

x( x - 21 ) = 10^2

x^2 - 21x = 100

x^2 - 21x - 100 = 0

( x - 25 ) ( x + 4 ) = 0

x - 25 = 0 or x + 4 = 0

x = 25 or x = - 4

x can't be -4 since x is used in log terms

finally x = 25
2010-03-31 6:52 pm
log(x - 21) = 2 - log(x)
log(x - 21) + log(x) = 2
log[x(x - 21)] = 2
x(x - 21) = 10^2
x^2 - 21x - 100 = 0
x^2 + 4x - 25x - 100 = 0
(x^2 + 4x) - (25x + 100) = 0
x(x + 4) - 25(x + 4) = 0
(x + 4)(x - 25) = 0

x + 4 = 0
x = -4

x - 25 = 0
x = 25

But...

log(-4) is invalid.

∴ x = 25
2010-03-31 6:09 pm
Log(x-21)+log(x)=2
ln(x-21)/ln(x)+ln(10)=2
ln(x-21)+ln(x)=2ln(10)
e^(ln(x-21)+ln(x))=e^2ln(10)
x(x-21)=100
x=25
2010-03-31 5:56 pm
log(x-21)=2-log(x
log(x-21) + log(x) = 2

log( (x-21) x) = 2
log( x^2-21x) = 2
x^2-21x = 10^2
x^2-21x = 100

x^2-21x-100 = 0
x^2-25x+4x-100 =0
x(x-25)+4(x-25)=0
(x-25)(x+4)=0

x=25 or x=4
2016-12-10 8:27 pm
log [ x ( x - 21 ) ] = 2 x ( x - 21 ) = 10 ² x ² - 21 x = one hundred x ² - 21 x - one hundred = 0 x = [ - b ± ? ( b ² - 4 a c ) ] / 2 a x = [ 21 ± ? ( 441 + 400 ) ] / 2 x = [ 21 ± ? ( 841 ) ] / 2 x = [ 21 ± 29 ] / 2 x = 25 is ideal answer
2010-03-31 5:59 pm
10^log(x-21)=10^2-log(x). the 10 and the log then cancel out and you are left with x-21=10^2-x which simplifies to x-21=100-x and x to both sides and the x on the right cancels out and the other becomes x^2 and you are left with (x^2)-21=100 add 21 to both sides and the 21 on the left cancels out and you are left with x^2=121 take the square-root of both sides and you are left with x=11. hope this helps :)


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