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+5x-21=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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