If #log 2=a# and #log 3 = b#, evaluate #log(sqrt60sqrt2)#?

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Answer 1 · 2018-06-13T19:20:40

# 3/2*a+1/2*b+1/2log5#.

Using the Usual Rules of #log#,

#log(sqrt60sqrt2)=logsqrt60+logsqrt2#

#=1/2log60+1/2log2#,

#=1/2log(2^2*3*5)+1/2log2#,

#=1/2[log2^2+log3+log5]+1/2log2#,

#=1/2[2log2+log3+log5]+1/2log2#,

#=(log2+1/2log2)+1/2log3+1/2log5#,

#=3/2log2+1/2log3+1/2log5#,

# rArr log(sqrt60sqrt2)=3/2*a+1/2*b+1/2log5#.