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.
exponentiating to solve equations: solve the eaquation, cheack for extraneous solutions
log5(x+4)+log5(x+1)=2
4 answers
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
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
i dnt get bwat u did after x^2+5x-21=0. that is were i keep gettin stuck on
http://en.wikipedia.org/wiki/Quadratic_equation
1 answer