SOLUTION: Solve for x. a.) log sub2(log sub4{log sub3[log sub2 x]}) = -1 b.) log sub2(x+4) + log sub2(x-2) - log sub2(x-6) = 0

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Solve for x. a.) log sub2(log sub4{log sub3[log sub2 x]}) = -1 b.) log sub2(x+4) + log sub2(x-2) - log sub2(x-6) = 0      Log On


   

Question 942780: Solve for x.
a.) log sub2(log sub4{log sub3[log sub2 x]}) = -1
b.) log sub2(x+4) + log sub2(x-2) - log sub2(x-6) = 0
Found 2 solutions by MathLover1, MathTherapy:
Answer by MathLover1(20854) About Me  (Show Source):
You can put this solution on YOUR website!
a.
log+%282%2C%28log%284%2C%28log%283%2C%28log+%282%2Cx%29%29+%29+%29+%29+%29%29+=+-1 ...change the base starting from the first log left

log+%28%28log%284%2C%28log%283%2C%28log+%282%2Cx%29%29+%29+%29+%29+%29%29%2Flog+%282%29+=+-1

log+%28%28log%284%2C%28log%283%2C%28log+%282%2Cx%29%29+%29+%29+%29+%29%29=+-1%2Alog+%282%29+

log+%28%28log%284%2C%28log%283%2C%28log+%282%2Cx%29%29+%29+%29+%29+%29%29=+log+%282%5E-1%29+


log+%28%28log%284%2C%28log%283%2C%28log+%282%2Cx%29%29+%29+%29+%29+%29%29=+log+%281%2F2%29+


log%284%2C%28log%283%2C%28log+%282%2Cx%29%29+%29+%29%29+=+%281%2F2%29+

log%28%28log%283%2C%28log+%282%2Cx%29%29+%29+%29%29+%2Flog%284%29=+%281%2F2%29+

log%28%28log%283%2C%28log+%282%2Cx%29%29+%29+%29%29+=+%281%2F2%29+log%284%29


log%28%28log%283%2C%28log+%282%2Cx%29%29+%29+%29%29+=+log%284%5E%281%2F2%29+%29


log%28%28log%283%2C%28log+%282%2Cx%29%29+%29%29+%29+=+log%28sqrt%284%29+%29

log%28%28log%283%2C%28log+%282%2Cx%29%29+%29%29+%29+=+log%282+%29

log%283%2C%28log+%282%2Cx%29%29+%29+=+2+

log%28%28log+%282%2Cx%29%29%29%2F+log%283%29=+2+

log%28%28log+%282%2Cx%29%29%29=+2log%283%29+

log%28%28log+%282%2Cx%29%29%29=+log%283%5E2%29+

log%28%28log+%282%2Cx%29%29%29=+log%289%29+

log+%282%2Cx%29=+9

log+%28x%29%2Flog+%282%29=+9

log+%28x%29=+9log+%282%29


log+%28x%29=+log+%282%5E9%29

x=+2%5E9

x=+512


b.)
log%282%2C%28x%2B4%29%29+%2B+log%282%2C%28x-2%29%29+-+log%282%2C%28x-6%29%29+=+0

log%282%2C%28x%2B4%29%28x-2%29%29+-+log%282%2C%28x-6%29%29+=+0

log%282%2C%28%28x%2B4%29%28x-2%29%29%2F%28x-6%29%29+=+0

log%28%28%28x%2B4%29%28x-2%29%29%2F%28x-6%29%29%2Flog%282%29+=+0

log%28%28%28x%2B4%29%28x-2%29%29%2F%28x-6%29%29+=+0%2Alog%282%29

log%28%28%28x%2B4%29%28x-2%29%29%2F%28x-6%29%29+=+0.................since log%2810%2C1%29=0, we have

log%28%28%28x%2B4%29%28x-2%29%29%2F%28x-6%29%29+=+log%2810%2C1%29 then

%28%28x%2B4%29%28x-2%29%29%2F%28x-6%29+=1

%28x%2B4%29%28x-2%29+=1%28x-6%29

x%5E2-2x%2B4x-8+=x-6

x%5E2%2B2x-8-x%2B6=0

x%5E2%2Bx-2=0

x%5E2-x%2B2x-2=0 ..factor

%28x%5E2-x%29%2B%282x-2%29=0

x%28x-1%29%2B2%28x-1%29=0

%28x%2B2%29%28x-1%29=0

solutions:
if %28x%2B2%29=0=>x=-2
if %28x-1%29=0=>x=1


Answer by MathTherapy(10704) About Me  (Show Source):
You can put this solution on YOUR website!
   a.) log sub2(log sub4{log sub3[log sub2 x]}) = -1
   b.) log sub2(x+4) + log sub2(x-2) - log sub2(x-6) = 0

a) PAINSTAKINGLY LONG solution by the other person.
   log+%282%2C+%28log+%284%2C+%28log+%283%2C+%28log+%282%2C+%28x%29%29%29%29%29%29%29%29+=+-+1

b) Other person's solutions are EXTRANEOUS, and therefore, UNACCEPTABLE!!!
   b.) log sub2(x+4) + log sub2(x-2) - log sub2(x-6) = 0
       log+%282%2C+%28x+%2B+4%29%29+%2B+log+%282%2C+%28x+-+2%29%29+-+log+%282%2C+%28x+-+6%29%29+=+0
       
        Smallest log: (x - 6), so x - 6 > 0. Therefore, x > 6.
        We then have: log+%282%2C+%28x+%2B+4%29%29+%2B+log+%282%2C+%28x+-+2%29%29+-+log+%282%2C+%28x+-+6%29%29+=+0, with x > 6.

The 2 solutions the other person got, x = - 2, and x = 1 are < 6, NOT > 6, obviously.
This makes them EXTRANEOUS, thereby leading to this equation having NO SOLUTIONS!