Given log4=0.6021, log9=0.9542, and log12=1.-792, how do you find log 108,000?

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Answer 1 · 2016-11-21T09:19:23

#log108000=5.0334#

#log108000=log(108xx1000)=log108+log1000#

= #3+log108#

= #3+log(4xx27)=3+log4+log27#

= #3+log4+log3^3#

= #3+log4+3log3#

Now we have #log4=0.6021# and as #log3^2=log9=0.9542#,

we have #2log3=0.9542# and hence #log3=0.9542/2=0.4771#

Hence #log108000=3+0.6021+3xx0.4771#

= #3.6021+1.4313#

= #5.0334#