logarithmic equation?

Question

Can anyone show me the steps in solving this?

log base 2 (3-x) + log base 2 (2-x) = log base 2 (1-x)

Thanks

Answer 1

we know that log basex a + log basex b = log basex ab
so log base 2 (3-x) + log base 2 (2-x)´= log base 2 (3-x)(2-x)
=log base 2 (6-5x+x^2)
but L.H.S = R.H.S
SO log base 2 (6-5x+x^2) = log base 2 (1-x)
as log on both sides get cancelled,
6-5x+x^2=1-x
x^2-4x+5=0
by using the formula for solving quadratic equations x=4+2i/2,4-2i/2=2+i,2-i
where i is the square root of -1

Answer 2

first add the logs on the left by multiplying the (3-x)(2-x)
log2(6 - 5x + x^2) = log 2 (1-x)
Property:
if log 2 m = log 2 n then m must equal n
Thus,
x^2 - 5x + 6 = 1 - x get this equal to 0 to solve the quadratic

Be sure to check you answers in the original equation to make sure that (3 - x) nor (2 - x) ends up negative.

Answer 3

when u have addition such as log base x (y) + log base x (z) its simply log base x (yz) so

1)log base2 (3-x)(2-x)-log base2 (1-x)=0

and when u subtract it will be y/z so

2)log base2 (3-x)(2-x)/(1-x)=0

now use he propriety : if log base x (b) =a then x^a =b

3)2^0 =(3-x)(2-x)/(1-x)
1= .....
cross multiply

4) 1-x= (3-x)(2-x)
1-x= 6-5x+x^2
x^2-4x-5=0
(x-5)(x+1)=0
x=-1, 5
x cannot be 5 because then u would have log of a negative number which doesnt exist so
x=-1 final answer

Answer 4

all log base 2, so, omit them...
6 - 2x + 2 - x = 1 - x
-2x = -7
x = 7/2

Answer 5

-------------------------------
( 2 as the base)
log a + log b = log m

log (a * b) =log m
log (a * b) - log m =0
log (a * b)/m =0

2^0 = (a* b)/m
1 = (a * b)/m
------------------------------------
do the same in ur Q u got

1 = [(3-x) (2-x)] / (1-x)

(1-x) =[(3-x) (2-x)]

1-x = 6 -3x-2x+x²
6 -3x-2x+x² -1 +x=0
x² -4x +5 =0
(x-5)(x+1)=0

so,x=(5) or (-1)

Answer 6

(3 - x)(2 - x) = 1 - x
6 - 5x + x² = 1 - x
x² - 4x + 5 = 0
x = [ 4 ±√(16 - 20) ] / 2
x = [ 4 ± √(- 4) ] / 2
x = [ 4 ± 2 i ] / 2
x = 2 ± i