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November 28, 2014

November 28, 2014

Posted by **james** on Monday, March 19, 2012 at 4:23am.

log3(x - 5) + log3(x + 3) = 2 <--- both log3.. 3 is subscript

log7x + log7(x - 1) = log72x <--- all log7.. 7 is subscript.

- math -
**Reiny**, Monday, March 19, 2012 at 8:40amuse your log rules:

log (AB) = logA + logB

and

log (A/B) = logA - logB

1st one:

log_{3}[(x-5)(x+3)] = 2

by definition of logs

(x-5)(x+3) = 3^2 = 9

x^2 - 2x -24=0

(x-6)(x+4) = 0

x = 6 or x=-4 , but x=-4 would make log_{3}(x+3) undefined, so

x = 6

2nd:

log_{7}(x) + log_{7}(x-1) - log_{7}(2x) = 0

log_{7}[x(x-1)/(2x)] = 0

log_{7}[(x-1)/2] = 0

(x-1)/2 = 7^0

(x-1)/2 = 1

x-1=2

x = 3

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