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gsingh7 · Jul 14, 2008, 02:09 PM

1) Solver for x:

log7(x+4) + log7(x-2)=1

ebaines · Jul 15, 2008, 05:37 AM

First, recall the rule: Log(a*b) = log(a) + log(b). So the first step in your problem is to combine the two log terms - here I assume that when you wrote log7(x+4) you meant:

$log _7 (x+4)$ , correct?

So:
$ log(x+4) + log(x-2) = 1\\ log((x+4)*(x-2)) = log(x^2 +2x-8) = 1 $

Now raise both sides by the power of 7 (the base of the log), recalling that
$ 7 ^ {log_7 a} = a \\ 7 ^ {log _7 (x^2 +2x -8)} = 1\\ x^2 +2x -8 = 7^1 = 7 $

Can you take it from here?

chongxiang · May 2, 2009, 05:17 AM

Originally Posted by gsingh7

1) Solver for x:

log7(x+4) + log7(x-2)=1

7x+4+7x-14=1
14x=1+14-4
14x=11
x=1.3