SOLUTION: Solve i) {{{log(base3)(2x + 1) = 2 + log (base3) (3x - 11)}}} ii) {{{log(base4)y + log (base2)y = 9 }}} *Please answer as soon as possible bro :)

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solve i) {{{log(base3)(2x + 1) = 2 + log (base3) (3x - 11)}}} ii) {{{log(base4)y + log (base2)y = 9 }}} *Please answer as soon as possible bro :)      Log On


   

Question 473615: Solve
i) log%28base3%29%282x+%2B+1%29+=+2+%2B+log+%28base3%29+%283x+-+11%29
ii) log%28base4%29y+%2B+log+%28base2%29y+=+9+

*Please answer as soon as possible bro :)
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
i) log(base3)(2x + 1) = 2 + log (base3) (3x - 11)
ii)log(base4)y + log (base2)y = 9
**
i) log3(2x+1)=2+log3(3x-11)
log3(2x+1)-log3(3x-11)=2
log3[(2x+1)/(3x-11)]=2
convert to exponential form: (base(3) raised to log of number(2)=number(2x+1)/(3x-11)
3^2=(2x+1)/(3x-11)=9
2x+1=27x-99
25x=100
x=4
Check:
log3(2x+1)=2+log3(3x-11)
log3(8+1)=2+log3(12-11)
log3(9)=2+log3(1)
2=2+0
..
ii)log4y+log2y=9
change to base 2
log2y/log2(4)+log2y=9
log2y/2+log2y=9
LCD:2
log2y+2log2y=18
3log2y=18
log2y=6
convert to exponential form: (base(2) raised to log of number(6)=number (y)
2^6=y
y=64
Check:
log4y+log2y
log4(64)+log2(64)=3+6=9