SOLUTION: Hi, i need to find the solution set of the equation : 2 log (base 2) (x+2) - log (base 2) (7x+2) = log (base x) x^2 - log (base square root of 3) 3 Please show the solution so

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Hi, i need to find the solution set of the equation : 2 log (base 2) (x+2) - log (base 2) (7x+2) = log (base x) x^2 - log (base square root of 3) 3 Please show the solution so       Log On


   

Question 864939: Hi, i need to find the solution set of the equation :
2 log (base 2) (x+2) - log (base 2) (7x+2) = log (base x) x^2 - log (base square root of 3) 3
Please show the solution so that i know what to do in the exam. Thanks.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

Re TY;  Good Catch and Good work!



Log%282%2C%28%28x%2B2%29%5E2%29%2F%287x%2B2%29%29+=+0

%28x%2B2%29%5E2%2F%287x%2B2%29+=+1  

 x^2 + 4x + 4 = 7x + 2

x^2 - 3x - 2 = 0

(x-2)(x-1) = 0,   x = 1, 2

and IF...:)

log%28x%2C%28x%5E2%29%29+=+2 and log%283%2C3%29+=+1

Therefore:

Log%282%2C%28%28x%2B2%29%5E2%29%2F%287x%2B2%29%29+=+2-1+=+1

%28x%2B2%29%5E2%2F%287x%2B2%29+=+2     |Note:  If log(2,X) = 1  then 2^1 = X

  x^2 + 4x + 4 = 14x + 4

  x^2 - 10x = 0

 x(x-10) = 0,   x = 10    |Note:  root x = 0 is an extraneous root