If log x = log 3 + 2 log 2- (3/4) log 16. the value of x is: a
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log x= log3 + 2log2 - (3/4) log(2)^4:(16=2^4)
so now
logx=log3 + 2log2 - (3/4)*4 log2
logx=log3 + 2log2 - 3log2
logx=log3 - log2
logx=log(3/2)
by taking antilog on both side we get
x=3/2
so now
logx=log3 + 2log2 - (3/4)*4 log2
logx=log3 + 2log2 - 3log2
logx=log3 - log2
logx=log(3/2)
by taking antilog on both side we get
x=3/2
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