**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: log_{2}(2 - x) = 1 - log_{2}(3 - x)

# Solution

First we could do a simple manouver to isolate the whole number.

**log**_{2}(2 - x) + log_{2}(3 - x) = 1

Following the Product Rule:

**log**_{2}(2 - x) (3 - x) = 1

Then we could rewrite the equation to its exponential form.

**21 = (2 - x) (3 - x)
**

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****2 = 6 - 5x + x**^{2} <-solving for x

**2 = x**^{2} - 5x + 6 <-rewrite the equation

**0 = x**^{2} - 5x + 6 - 2 <-subtract 2 from both sides to have a quadratic equation.

**0 = x**^{2} - 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

**log**_{2}(2 - 4) = 1 - log_{2}(3 - 4)

**log**_{2}(-2) = 1 - log_{2}(-1)

**x = 4 is rejected.**

For x = 1

**log**_{2}(2 - 1) = 1 - log_{2}(3 - 1)

**log**_{2}(1) = 1 - log_{2}(2)

**x = 1 is accepted.**

**x = 1 is the answer.**

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