solve the following equation.
2log x = log 2x
2logx=log2x2log x = log 2x2logx=log2x
logx2=log2xlog x^2= log 2xlogx2=log2x since (xlogy=logyx)(xlogy= logy^x)(xlogy=logyx)
Since the bases of logarithm are the same, set the arguments equal.
x2=2xx^2=2xx2=2x
x2−2x=0x^2-2x=0x2−2x=0
x(x−2)=0x(x-2)=0x(x−2)=0
x=0x=0x=0
x−2=0x-2=0x−2=0
x=2>0.x=2>0.x=2>0.
Answer: x=2.x=2.x=2.
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