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
|
|
|
