log2(2 - x) = 1 - log2(3 - x); Solve for X?

Question

Show your work. =)

Basics of logarithms... you remember those, don't you?

Answer 1

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

log(2)(2 - x) + log(2)(3 - x) = 1
log(2)((2 - x)(3 - x)) = 1
log(2)(6 - 2x - 3x + x^2) = 1
log(2)(x^2 - 5x + 6) = 1

x^2 - 5x + 6 = 2^1
x^2 - 5x + 6 = 2
x^2 - 5x + 4 = 0
(x - 4)(x - 1) = 0
x = 4 or 1

since plugging in 4 for x would get you a negative log

ANS : x = 1

Answer 2

log2(2-x) = 1 - log2(3-x)
To solve this you need to know the rules of logs and a little algebra.
First note that the base is 2 for both of the log terms so we can perform the necessary manipulations.
Now you must get both these terms involving logs on one side so add log2(3-x) to both sides of the equation.
log2(2-x) + log2(3-x) = 1 - log2(3-x) + log2(3-x)
which gives
log2(2-x) + log2(3-x) = 1
now recall the rule which states "the sum of the logs = the log of the product" written
loga(x) + loga(y) = loga(xy).
so your problem becomes
log2[(2-x)*(3-x)] = 1
Then apply the rule which transforms a log problem into an exponential problem. That is l
oga(x) = y can be written a^y=x.
Now your problem can be written:

2^1 = (2-x)*(3-x) or (2-x)*(3-x) =2

now expand the left side to get

6 - 5x + x^2 = 2 which can be written

x^2 - 5x + 6 = 2 now subtract 2 from both sides.

x^2 - 5x + 6 - 2 = 2 - 2

x^2 - 5x + 4 = 0

now factor this equation
x^2 - 4x - x + 4 = 0
x(x - 4) - (x - 4) = 0
(x - 4)(x - 1) = 0

so the two solutions are 1 and 4 HOWEVER THEY ARE NOT BOTH CORRECT.

if you substitute 4 in the original equation it becomes
log2(-2) = 1 - log2(-1)
which cannot be computed since we cannot take the log of a negative number.

if we substitute 1 in the original equation it becomes
log2(1) = 1 - log2(2)
0 = 1 - 1 = 0

so the only correct solution is x = 1

remember log 1 to any base is 0 [ log2(1) the left hand side]
and log of any number to the same base is always 1 [log2(2) on the right hand side]

Answer 3

I assume log2(2 - x) means the log, base 2, of the quantity 2 - x.

log2(2 - x) = 1 - log2(3 - x)
Move the log to the other side and get
log2(2 - x) + log2(3 - x) = 1
Use a log law to combine and get
log2((2 - x)(3 - x)) = 1
So, the log, base 2, of the mess is one, which means
2^1 = (2 - x)(3 - x)
or, simplifying
2 = x^2 - 5x + 6
or
x^2 - 5x + 4 = 0
so that we get
(x - 4)(x - 1) = 0
or that x = 4 or x = 1.

Answer 4

I take it you mean the logs with base 2.

log_2(2 - x) = 1 - log_2(3 - x) (Transpose).
log_2(2 - x) + log_2(3 - x) = 1 (Rule of logs, addition becomes multiplication).
log_2[(2 - x)(3 - x)] = 1
log_2(x² - 5x + 6) = 1

Understand what logs mean. It's the value of the power that must be put onto the base, to give the given number.
So here, the base is 2, and the number is (x² - 5x + 6).
We are told that this equals 1 - the equation about.
This implies:
x² - 5x + 6 = 2^1
x² - 5x + 6 = 2
x² - 5x + 4 = 0

This being a quadratic equation, you can use the formula to identify the roots:
[-b ± √(b² - 4.a.c)] / 2a
So x = 4 or x = 1.
Now back to the original equation:
log_2(2 - x) = 1 - log_2(3 - x) & subsititute values in.
For x = 4 you get minus log - no good.
For x = 1 you get a matching answer.

log_2(2 - x) = 1 - log_2(3 - x)
log_2(2 - 1) = 1 - log_2(3 - 1)
log_2(1) = 1 - log_2(2)
So what power on base 2 gives 1? Answer is 0.
And what power on base 2 give 2? Answer is 1.

log_2(1) = 1 - log_2(2)
0 = 1 - 1
0 = 0

Answer 5

Yes, we do remember logarithms, why don't you? I assume that log2 means log base 2. Then

log2(2 - x) = 1 - log2(3 - x)
2-x=2/(3-x)
(2-x)(3-x)=2
x²-5x+6=2
x²-5x+4=0
(x-4)(x-1)=0
x=1 or x=4

Edit: fixed minor spelling error. And doug: log a + log b = log (ab), not log (a+b).

Answer 6

log2(2 - x) + log2(3 - x)=1
log2{(2-x)(3-x)} = log2(2)
(Because log of any nmuber having same base is one)
Remove log both side
(2-x)(3-x) =2
6 -2x -3x + x^2 = 2
x^2 -5x +4 =0
x^2 -4x -x +4=0
x(x-4) -1(x-4) =0
(x-1)(x-4)=0
Therefore x=4 and x=1

Answer 7

If you remembered basics of logs, you could work this problem for yourself

log2(2-x)+log2(3-x)=1
2^(5-2x) = 2^1 = 2 ==> 5-2x=1 so x=2

If you want to do the arithmetic, remember that
loga(x) = ln(x)/ln(a)


Doug

Answer 8

thanks for the 2 points.... remember what...?

log(a+b)= loga +logb....

BTW, 2 log2 is 0.6