Posted by Kayla on Sunday, March 20, 2011 at 8:42pm.
exponentiating to solve equations: solve the eaquation, cheack for extraneous solutions
log5(x+4)+log5(x+1)=2

acc. alg 2  bobpursley, Sunday, March 20, 2011 at 8:45pm
remember log a + log b= log ab
do that, then take the antilog..
log5 (x+4)(x+1)=2
(x+4)(x+1)=25
then proceed to solve.

acc. alg 2  Damon, Sunday, March 20, 2011 at 8:52pm
log5 [(x+4)(x+1)] = 2
5^log5 [(x+4)(x+1)] = 5^2 = 25
(x+4)(x+1) = 25
x^2 + 5 x + 4 = 25
x^2 + 5 x  21 = 0
x = [5 +/ sqrt(25+84) ]/2
= [5 +/ sqrt(109)]/2
= [ 2.5 +/ 5.22 ]
= 2.72 or  7.72
You can not take log of a negative number so use 2.72

acc. alg 2  Kayla, Sunday, March 20, 2011 at 9:10pm
i dnt get bwat u did after x^2+5x21=0. that is were i keep gettin stuck on

acc. alg 2  Damon, Monday, March 21, 2011 at 9:13am
http://en.wikipedia.org/wiki/Quadratic_equation
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