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Page history last edited by PBworks 13 years, 10 months ago

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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