Log Equation

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mathxyz

Junior Member
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Jul 8, 2005
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112
I came across this question.

log_a (x-1) - log_a (x+6) = log_a (x-2) - log_a (x + 3)

Do I equate everything to zero?

MY WORK:

log_a (x-1) - log_a (x+6) - log_a (x-2) - log_a (x + 3) = 0

log_a (x-1)/(x+6) - log_a (x-2)/(x+3) = 0

Am I right so far? If not, how should I attack this question?

My answer: x = 3/2

Book's answer: x = 9/2
 
log_a (x-1) - log_a (x+6) = log_a (x-2) - log_a (x + 3)
Click to expand...

Rather than equating everything to zero in the first step try this

log_a(x-1)/(x+6)=log_a(x-2)/(x+3)

Since log_a is common to both,it can be removed without affecting the equation.

(x-1)/(x+6)=(x-2)/(x+3)

Solve this!
 
That's okay, but I would lean toward simplifying first.
log_a ((x-1)/ (x+6)) = log_a ((x-2) / (x + 3))
a^((x-1)/ (x+6)) = a^((x-2) / (x + 3))
(x-1)/ (x+6) = (x-2) / (x + 3)
(x-1)(x+3) = (x-2)(x+6)
x^2+2x-3 = x^2+4x-12
Then gather like terms.
9 = 2x
x = 9/2

You were right as far as you showed your work except for the sign on
-log_a (x + 3)
It should be + after you move it.

P.S. Sorry, askmemath. Didn't see you till I had hit submit.
 
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