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Grade: 12th pass
        
please can you help me by solving this question log2x+3(logx-logy)=log(2x4/y3)
one year ago

Answers : (2)

Rupesh Ojha
23 Points
							
Hello,
log 2x + 3log x - 3log y= log 2x⁴ - log y³
(Solving brackets and using log m/n = log m - log n)
log 2x + 3 log x - 3log y= 4log 2x - 3log y
(Log(m^n) = nlogm)
log 2x + 3log x - 3log y = 4log 2x -3log y
3 log 2x = 3 log x
log 2x = log x
 
I think you could carry on from here. Hope it helped!
one year ago
shankar poudel
14 Points
							
log2x+3(logx-logy)=log(2x4/y3)
Lhs=log2x+3(logx-logy)
     =  log2x+3logx-3logy
      =log2x+logx3-logy3     {logxm=mlogx}
      =log(2x.x3)-logy3  {log(xy)=logx+logy}
      =log2x4-logy3        {log(x/y)=logx-logy}
     =log(2x4/logy3)
one year ago
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