How do you evaluate #\log \sqrt{7\times 5\times 2}#?

2 Answers
Rhys
Dec 10, 2017

# = 1/2 ( 1 + log 7 ) approx 0.923 #

Explanation:

We must use our log laws:

# alpha log beta = log beta ^ alpha #
#log gamma + log delta = log gamma delta #
#log_alpha alpha = 1 #

So hence:

#log sqrt( 7*5*2) = 1/2 log 7*5*2 #

#=>1/2 log 7 * 10 #

Now using another one of our laws:

#=> 1/2( log 7 + log 10 )#

#=> 1/2 ( log 7+ 1 )#

# approx 0.923#

George C.
Dec 10, 2017

#log sqrt(7 xx 5 xx 2) = 1/2 log 7 + 1/2 ~~ 0.922549#

Explanation:

Note that:

#log sqrt(7xx5xx2) = log (7xx10)^(1/2)#

#color(white)(log sqrt(7xx5xx2)) = 1/2 log (7xx10)#

#color(white)(log sqrt(7xx5xx2)) = 1/2 (log 7 + log 10)#

#color(white)(log sqrt(7xx5xx2)) = 1/2 (log 7 + 1)#

#color(white)(log sqrt(7xx5xx2)) = 1/2 log 7 + 1/2#

What about #log 7# ?

We could go to a log table or calculator to find:

#log 7 ~~ 0.84509804#

Suppose we do not have access to that, but do know a few standard values:

#ln(10) ~~ 2.302585093#

#log(2) ~~ 0.30103#

and we know:

#ln(1+t) = t - t^2/2 + t^3/3 - t^4/4 +...#

Then note that:

#2 log 7 = log 7^2#

#color(white)(2 log 7) = log 49#

#color(white)(2 log 7) = log (100/2 * 49/50)#

#color(white)(2 log 7) = log 100 - log 2 + log (1 - 1/50)#

#color(white)(2 log 7) = 2 - log 2 + ln (1 - 1/50) / ln 10#

#color(white)(2 log 7) = 2 - log 2 - 1/ln 10 (1/50+1/(2 * 50^2) + 1/(3 * 50^3)+...)#

#color(white)(2 log 7) ~~ 2 - 0.30103 - 1/2.302585093 (1/50+1/5000 + 1/375000) ~~ 1.690196#

So:

#1/2 log 7+1/2 ~~ 1.690196/4+0.5 = 0.922549#

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