SOLUTION: I need help solving these problems. {{{ 2log(3,x)-log(3,x-4)=(2)+log(3,2)}}} and {{{(1/2)log (4x+5)= log x}}}

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I need help solving these problems. {{{ 2log(3,x)-log(3,x-4)=(2)+log(3,2)}}} and {{{(1/2)log (4x+5)= log x}}}      Log On


   

Question 458375: I need help solving these problems.
+2log%283%2Cx%29-log%283%2Cx-4%29=%282%29%2Blog%283%2C2%29
and
%281%2F2%29log+%284x%2B5%29=+log+x
Found 2 solutions by josmiceli, lwsshak3:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Use the rule +log%28a%29+%2B+log%28b%29+=+log%28a%2Ab%29+
and the rule +log%28a%29+-+log%28b%29+=+log%28a%2Fb%29+
and the rule +a%2Alog%28b%29+=+log%28b%5Ea%29+
----------------------------------------
+2log%283%2Cx%29+-+log%283%2Cx-4%29+=+%282%29+%2B+log%283%2C2%29
+log%283%2Cx%5E2%29+-+log%283%2Cx-4%29+=+log%283%2C9%29+%2B+log%283%2C2%29+
+log%283%2C+x%5E2+%2F+%28x-4%29+%29+=+log%283%2C+9%2A2+%29+
+x%5E2+%2F+%28x-4%29+=+18+
+x%5E2+=+18%2A%28x-4%29+
+x%5E2+=+18x+-+72+
+x%5E2+-+18x+%2B+72+=+0+
Use quadratic formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+a+=+1+
+b+=+-18+
+c+=+72+
x+=+%28-%28-18%29+%2B-+sqrt%28+%28-18%29%5E2+-+4%2A1%2A72+%29%29%2F%282%2A1%29+
x+=+18+%2B-+sqrt%28+324+-+288+%29%29+%2F+2+
x+=+18+%2B-+sqrt%28+36+%29%29+%2F+2+
+x+=+%2818+%2B+6%29%2F2+
+x+=+12+
and
+x+=+%28+18+-+6+%29%2F2+
+x+=+6+
check:
+%28x-6%29%2A%28x-12%29+=+x%5E2+-+18x+%2B+72+
OK
---------
%281%2F2%29log+%284x%2B5%29=+log+x
+log%28+sqrt%284x%2B5%29%29+=+log%28x%29+
+sqrt%28+4x+%2B+5+%29+=+x+
Square both sides
+4x+%2B+5+=+x%5E2+
+x%5E2+-+4x+-+5+=+0+
Complete the square:
+x%5E2+-+4x+=+5+
+x%5E2+-+4x+%2B+%28%28-4%29%2F2%29%5E2+=+5+%2B+%28-4%2F2%29%5E2+
+x%5E2+-+4x+%2B+4+=+5+%2B+4+
+x%5E2+-+4x+%2B+4+=+9+
+%28x-2%29%5E2+=+3%5E2+
+x-2+=+3+
+x+=+5+
also
+x+-+2+=+-3+
+x+=+-1+

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
2log3(x)-log3(x-4)=2+log3(2)
(1/2)log (4x+5)= log x
..
2log3(x)-log3(x-4)=2+log3(2)
2log3(x)-log3(x-4)-log3(2)=2
2log3(x)-(log3(x-4)+log3(2))=2
place under single log
log3[x^2/(x-4)(2)]=2
convert to exponential form (base(3) raised to log of number(2)=number[x^2/(x-4)(2)]
3^2=x^2/(x-4)(2)=9
x^2/(2x-8)=9
x^2=18x-72
x^2-18x+72=0
(x-9)(x-8)=0
x=9
or
x=8
...
(1/2)log (4x+5)= log x
(1/2)log (4x+5)- log x=0
log[(4x+5)^1/2/x]=0
10^0=(4x+5)^1/2/x=1
(4x+5)^1/2=x
square both sides
4x+5=x^2
x^2-4x-5=0
(x-5)(x+1)=0
x=5
or
x=-1 (reject, ( x>0)