Grade 12th passAlgebra
please can you help me by solving this question log2x+3(logx-logy)=log(2x4/y3)
please can you help me by solving this question log2x+3(logx-logy)=log(2x4/y3)
2 Answers
Rupesh Ojha
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!
shankar poudel
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)
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