SOLUTION: Please help me solve this equation: log16x + log8 + log2x = 19/12

Question 492142: Please help me solve this equation: log16x + log8 + log2x = 19/12
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
log16x + log8 + log2x = 19/12
**
log16x + log8 + log2x = 19/12
place under a single log
log[16x*8*2x]=19/12
convert to exponential form: base(10) raised to log of number(19/12)=number[16x*8*2x]
10^(19/12)=16x*8*2x=256x^2
10^(19/12)=256x^2
x^2=256/10^(19/12)=38.3119/256=.1497
x=�√.1497
x=.3869
or
x=-.3869 (reject, x>0)
Check:
log(16*.3869)+log(8)+log(2*.3869)=1.5834
19/12=1.5833
RELATED QUESTIONS
Please help me solve this problem: log16x+log4x+log2x=7. I know you have to use the... (answered by lwsshak3)

Please solve:... (answered by Alan3354,stanbon)

Please help me solve this equation: 2 log8 5+4log8 a - 5 log8 b... (answered by ewatrrr)

Please Help Me Solve This Equation..Log8 x~4 log8... (answered by fcabanski)

Please help me solve log2x + logx =... (answered by stanbon)

Solve the equation: log2x=6. Help... (answered by Earlsdon)

log8 (32) please help me solve (answered by jsmallt9)

please help me solve this equation: rewrite 2(log8)+log4 as the logarithum of a single... (answered by user_dude2008)

Please help me solve the following equation: log2x+2(2x^2=8x+6)=2 (answered by richwmiller)