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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 3, 2013 by bradely

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