SOLUTION: Determine the real values of x such that {{{log(5x+9,(x^2+6x+9))+log(x+3,(5x^2+24x+27))=4}}}

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Determine the real values of x such that {{{log(5x+9,(x^2+6x+9))+log(x+3,(5x^2+24x+27))=4}}}      Log On


   

Question 1114883: Determine the real values of x such that log%285x%2B9%2C%28x%5E2%2B6x%2B9%29%29%2Blog%28x%2B3%2C%285x%5E2%2B24x%2B27%29%29=4
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


log%285x%2B9%2C%28x%5E2%2B6x%2B9%29%29%2Blog%28x%2B3%2C%285x%5E2%2B24x%2B27%29%29=4

log%285x%2B9%2C%28%28x%2B3%29%5E2%29%29%2Blog%28x%2B3%2C%28%28x%2B3%29%285x%2B9%29%29%29=4



log%285x%2B9%2C%28%28x%2B3%29%5E2%29%29%2B1%2Blog%28%28x%2B3%29%2C%285x%2B9%29%29=4

log%285x%2B9%2C%28%28x%2B3%29%5E2%29%29%2Blog%28%28x%2B3%29%2C%285x%2B9%29%29=3

2%2Alog%285x%2B9%2C%28x%2B3%29%29%2Blog%28%28x%2B3%29%2C%285x%2B9%29%29=3

Make a u substitution to make the notation easier....

u+=+log%285x%2B9%2Cx%2B3%29

Then

1%2Fu+=+log%28x%2B3%2C5x%2B9%29

Then the equation is

2u+%2B+1%2Fu+=+3
2u%5E2%2B1+=+3u
2u%5E2-3u%2B1+=+0
%282u-1%29%28u-1%29+=+0
u+=+1%2F2 or u+=+1

(1) log%285x%2B9%2Cx%2B3%29+=+1%2F2
or
(2) log%285x%2B9%2Cx%2B3%29=1

See where each of those solutions leads us....

(1) %285x%2B9%29%5E%281%2F2%29+=+x%2B3
5x%2B9+=+x%5E2%2B6x%2B9
x%5E2%2Bx+=+0
x%28x%2B1%29+=+0
x+=+0 or x+=+-1

Check x=0: x+3 = 3; 5x+9 = 9
log%289%2C9%29%2Blog%283%2C%283%2A9%29%29+=+1%2B3+=+4 CHECK!

Check x=-1: x+3 = 2; 5x+9 = 4
log%284%2C4%29%2Blog%282%282%2A4%29%29+=+1%2B3+=+4 CHECK!

(2) %285x%2B9%29%5E1+=+x%2B3
5x%2B9+=+x%2B3
4x+=+-6
x+=+-3%2F2

Check x = -3/2: x+3 = 3/2; 5x+9 = 3/2
CHECK!

Answer: There are 3 solutions: x=0, x=-1, and x=-3/2.