Everything you see below (except for the original problem) was written by a student. If they could do it, so can you! ;-)
Problem 1
Solve for x without using a calculator: log2(2 - x) = 1 - log2(3 - x)
Solution
First we could do a simple manouver to isolate the whole number.
log2(2 - x) + log2(3 - x) = 1
Following the Product Rule:
log2(2 - x) (3 - x) = 1
Then we could rewrite the equation to its exponential form.
21 = (2 - x) (3 - x)
2 = 6 - 5x + x2 <-solving for x
2 = x2 - 5x + 6 <-rewrite the equation
0 = x2 - 5x + 6 - 2 <-subtract 2 from both sides to have a quadratic equation.
0 = x2 - 5x + 4
0 = (x - 4) (x - 1) <-it factors nicely
We have two answers for x.
x - 4 = 0 and x - 1 = 0
x = 4 x = 1
Now, substitute x = 4 and x = 1 to check wether we accept or reject the answer. Since a logarithm is an exponent, we could never have an answer that will make the argument a negative number or zero. An any base raised to an any given exponent would never result to a negative power or a power that is equal to zero.
For x = 4
log2(2 - 4) = 1 - log2(3 - 4)
log2(-2) = 1 - log2(-1)
x = 4 is rejected.
For x = 1
log2(2 - 1) = 1 - log2(3 - 1)
log2(1) = 1 - log2(2)
x = 1 is accepted.
x = 1 is the answer.
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