How do you evaluate #log 1500#?

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Answer 1 · 2015-09-09T20:53:44

#2log2+log3+3log5#

Well we have

#log1500=log(15*100)=log15+log100#

#=log(3*5)+log10^2#

#=log3+log5+2log10#

#=2log10+log3+log5#

#=2log(2*5)+log3+log5#

#=2log2+log3+3log5#

Answer 2 · 2015-09-10T16:08:47

If #log# here is common log (base 10), then #log 1500 = 2+log15#

If #log 1500# means #log_10 1500#, then we have:

#log 1500 = log(100*15)#

# = log100 + log 15#

# = log10^2 + log 15#

# = 2 + log 15#

If you are using tables of logarithms, you'll need scientific notation:

#log 1500 = log(1.5 * 10^3)#

# = 3 + log(1.5)#

# = 3 + 0.1761# (from table)

# = 3.1761#