SOLUTION: Solve: {{{log(base16)(3x -1)}}} = {{{log(base4)(3x)}}}+ {{{log(base4)(0.5)}}}

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solve: {{{log(base16)(3x -1)}}} = {{{log(base4)(3x)}}}+ {{{log(base4)(0.5)}}}      Log On


   

Question 457002: Solve: log%28base16%29%283x+-1%29 = log%28base4%29%283x%29+ log%28base4%29%280.5%29
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
log%28base16%29%283x+-1%29 = log%28base4%29%283x%29+ log%28base4%29%280.5%29
..
log16(3x-1)=log4(3x)+log4(0.5)
change of base from 16 to 4 on left side of equation
log4(3x-1)/log4(16)
log4(3x-1)/4
..
log4(3x-1)/4=log4(3x)+log4(0.5)
log4(3x-1)/4=log4(3x*.5=log4(3x/2)
log4(3x-1)=4log4(3x/2)
log4(3x-1)-4log4(3x/2)=0
log4[(3x-1)/(3x/2)^4]=0
convert to exponential form: (base(4) raised to log of number(0)=number,[(3x-1)/(3x/2)^4]
4^0=[(3x-1)/(3x/2)^4]=1
(3x-1)=(3x/2)^4]=81x^4/16
16(3x-1)=81x^448x-16=81x^4
81x^4-48x+16=0
I don't know how to solve this algebriacally, but I got answers from my graphical program on my computer.
x=.362459
or x=.66667