SOLUTION: Log(base 2) (x+14)+log(base 2)x=5

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Log(base 2) (x+14)+log(base 2)x=5      Log On


   

Question 933453: Log(base 2) (x+14)+log(base 2)x=5
Found 2 solutions by MathLover1, MathTherapy:
Answer by MathLover1(20854) About Me  (Show Source):
You can put this solution on YOUR website!
log%282%2C%28x%2B14%29%29%2Blog%282%2Cx%29=5
log%282%2C%28%28x%2B14%29%2Ax%29%29=5
log%282%2C%28x%5E2%2B14x%29%29=5...in base 10
log%28%28x%5E2%2B14x%29%29%2Flog%28%282%29%29=5
log%28%28x%5E2%2B14x%29%29=5log%28%282%29%29
log%28%28x%5E2%2B14x%29%29=log%28%282%5E5%29%29
log%28%28x%5E2%2B14x%29%29=log%28%2832%29%29...since log same, then
x%5E2%2B14x=32...solve for x
x%5E2%2B14x-32=0...write 14x as -2x%2B16x
x%5E2-2x%2B16x-32=0...group
%28x%5E2-2x%29%2B%2816x-32%29=0
x%28x-2%29%2B16%28x-2%29=0
%28x-2%29%28x%2B16%29+=+0
solutions:
if %28x-2%29+=+0 then x=2
if %28x%2B16%29+=+0 then x=-16...we cannot use this solution
so, your solution is: highlight%28x=2%29


Answer by MathTherapy(10699) About Me  (Show Source):
You can put this solution on YOUR website!
Log(base 2) (x+14)+log(base 2)x=5

log+%282%2C+%28x+%2B+14%29%29+%2B+log+%282%2C+%28x%29%29+=+5
Smaller variable-expression, x, MUST be > 0. We then get: x MUST be > 0. The problem then becomes:
log+%282%2C+%28x+%2B+14%29%29+%2B+log+%282%2C+%28x%29%29+=+5, with x > 0.
       log+%282%2C+%28x+%2B+14%29%29x+=+5 ---- Applying log+%28a%2C+%28b%29%29+%2B+log+%28a%2C+%28c%29%29 = log+%28a%2C+%28bc%29%29
             x%28x+%2B+14%29+=+2%5E5 --- Converting to EXPONENTIAL form
             x%5E2+%2B+14x+=+32
         x%5E2+%2B+14x+-+32+=+0
     (x - 2)(x + 16) = 0 ----- Factoring TRINOMIAL
      x - 2 = 0      OR      x + 16 = 0 --- Setting each factor = to 0
         x = 2      OR           x = - 16 (IGNORE).

x = 2 is the ONLY solution, since the other value, - 16, is NOT > 0.