Solve: log2(x-3) + log2(x-2)=1

Archive SAT Subject Test Preparation
1

<p>I can solve and got 1 and 4 answers.
The solution key says only 4 because:
when you take 1, the equation becomes log2(-2) + log2(-1) = 1 ; but log of negative numbers is not possible. may be.</p>

<p>But
if you simplify,
or, log2(x-3) + log2(x-2)=1
or, log2{(x-3)(x-2)}=1</p>

<p>and now substitute x with 1 you can simplify and 1 satisfies the equation;
e.g. log2{(1-3)(1-2)}=1
or log2(2)=1 (True)</p>

<p>So My question is Is 1 a solution or not?</p>

2

<p>no…it’s really not .every element every item should be meaningful …if not.then the equation itself is meaningless…</p>

3

<p>Would you bother to elaborate you remark I didnot get what you mean.
I mean what is not meaningful?
BTW the 2 in log2(…) is base.</p>

4

<p>that is to say the rule only works when the number is always positive…although the answer 1 satisfies the simplified equation,but it doesn’t satisfy the original .so it is not an answer but an extraneous root</p>

5

<p>thanks.
. . . .</p>

6

<p>anytime (^ ^)</p>

7

<p>If it’s log base 2 (x^2 -5x +6 ) = 1
2^1 = x^2 - 5x + 6
0 = x^2 -5x +4
0 = (x-4) (x-1)
x = 4 or 1</p>