How do you approximate #log_5 50# given #log_5 2=0.4307# and #log_5 3=0.6826#?

2 Answers
Dec 21, 2016

Answer:

#log_5 50 ~= 2.4307#

Explanation:

#log_5 50# can be written as #log_ 5 (2 xx 25)#, which can be rewritten as #log_a(n xxm) = log_a n + log_a m#.

#=> log_5 2 + log_5 25#

We know the value of #log_5 2# is #0.4307# but to find the value of #log_5 25# we must use the change of base rule such that #log_a n =logn/loga#.

#=> 0.4307 + log25/log5#

Rewrite #25# and #5# in powers of #5#.

#=> 0.4307 + log(5^2)/log(5^1)#

We now use the power rule that #loga^n = nloga#.

#=> 0.4307 + (2log5)/(1log(5))#

#=> 0.4307 + 2#

#=> 2.4307#

Hopefully this helps!

Dec 21, 2016

Answer:

The question is what I call 'a little bit wooly'!

#log_5 50 ~~2.4054# with 1.04% error

Explanation:

The inclusion of the word 'approximate' implies a rather wider range of precision than is normally generated by standard calculations. There is a problem hear in as much as 'approximate' is rather open to interpretation. I bit like "how long is a piece of string"!
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(red)("Assumption: we are meant to use both of "log_5 3" and "log_5 2 #

Target is #" "log_5(50)#

#50 -> 3xx2^4 = 48# as an approximation of 50

Implying that

#log_5 50 " is somewhere near to " log_5 3+log_5 2^4#

#=>log_5 3+4log_5 2#

#=>0.6826+(4xx0.4307)" "=" "2.4054#

#log_5 50 ~~2.4054#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Actual value of #" "log_5 50 =(log_10 50)/(log_10 5) = 2.4307" to 4dp"#

Level of error# -> 2.4307-2.4054 = 0.0253+-0.00005#

= say 1.04%
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#log_5 50 ~~2.4054# with 1.04% error