Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,435 questions

17,804 answers

1,438 comments

774,966 users

Solve the logarithmic equation

+3 votes

.2LOG5(x-3) - LOG5(2X+1)=1

asked Dec 26, 2012 in PRECALCULUS by linda Scholar

2 Answers

+1 vote

2LOG5(x-3) - LOG5(2X+1)=1

Remember m log a =log a^m

LOG5(x-3)^2 - LOG5(2X+1)=1

Remember log a - log b = log (a/b) in any base.

So LOG5(x-3)^2 - LOG5(2X+1)= log5((x+3)^2/(2x+1)) = 1

Take each 5 to each sides' power

(x-3)^2/(2x+1) = 5^1 = 5

(x-3)^2 = 5 (2x+1)

Using formula ( a -b ) ^ 2 = a^2 + b^2 - 2ab

x^2 + 3^2 -2 (x) 3 = 5 (2x + 1)

Distributive property a ( b + c ) = ab +ac

x^2 + 3^2 -2 (x) 3 = 5 (2x)  + 5

x^2 +9 -6x = 10x  + 5

x^2 +9 -6x - 10x -  5 = 0

x^2 -16x + 4 = 0

Using formula x = ( -b ± sqrt ( b^2-4ac) )/ 2a

x =( -(-16) ± sqrt ( 256 - 16)) / 2

x = (16 ± sqrt 240) / 2

x = 8 ± sqrt 60

x = 8 + sqrt 60 and 8 - sqrt 60

 

 

 

 

answered Dec 26, 2012 by ashokavf Scholar
reshown Dec 26, 2012 by moderator
+3 votes

2LOG5(x-3) - LOG5(2X+1)=1

Remember m log a =log a^m

LOG5(x-3)^2 - LOG5(2X+1 = 1

Remember log a - log b = log (a/b) in any base.

So LOG5(x-3)^2 - LOG5(2X+1)= log5((x-3)^2/(2x+1)) = 1

Take each 5 to each sides' power

(x-3)^2/(2x+1) = 5^1 = 5

(x-3)^2 = 5 (2x+1)

Using formula ( a -b ) ^ 2 = a^2 + b^2 - 2ab

x^2 + 3^2 -2 (x) 3 = 5 (2x + 1)

Distributive property a ( b + c ) = ab +ac

x^2 + 3^2 -2 (x) 3 = 5 (2x)  + 5

x^2 +9 -6x = 10x  + 5

x^2 +9 -6x - 10x -  5 = 0

x^2 -16x + 4 = 0

Using formula x = ( -b ± sqrt ( b^2-4ac) )/ 2a

x =( -(-16) ± sqrt ( 256 - 16)) / 2

x = (16 ± sqrt 240) / 2

x = 8 ± sqrt 60

x = 8 + sqrt 60 and 8 - sqrt 60

x = 8 + 4sqrt 15 and 8 - 4sqrt 15

answered Jan 3, 2013 by peterson Rookie
reshown Jan 4, 2013 by bradely

Related questions

asked Sep 20, 2018 in PRECALCULUS by anonymous
asked Nov 1, 2014 in PRECALCULUS by anonymous
asked Nov 13, 2014 in CALCULUS by anonymous
...