How do you approximate #log_5 (10/9)# given #log_5 2=0.4307# and #log_5 3=0.6826#?

2 Answers
Advik S.
Mar 3, 2018

Approximately #0.0655#

Explanation:

The Logarithmic Division Rule states that:

#log_b(x/y)=log_b(x)-log_b(y)#

The Logarithmic Multiplication Rule states that:

#log_b(x*y)=log_b(x)+lob_b(y)#

We can apply the logarithmic division rule, so:

#log_5(10/9)#

becomes:

#log_5(10)-log_5(9)#

We can simplify the numbers in the brackets into the products of prime numbers:

#->log_5(2*5)-log_5(3*3)#

Now, we can apply the logarithmic multiplication rule, so:

#->(log_5(2)+log_5(5))-(log_5(3)+log_5(3))#

Now, we can substitute in the values given to us:

#->(0.4307+1)-(0.6826+0.6826)#

#(log_b(b)=1)#, always

#=(1.4307)-(1.3652)#

#=0.0655#

A08
Mar 3, 2018

We apply

  1. The Logarithmic Division Rule which states that

    #log(x/y)=logx-logy#

  2. The Logarithmic Multiplication Rule

    #log(x*y)=logx+logy#

  3. And the Logarithmic Power Rule

    #log(x^y)=ylogx#

Given number can be written as

#log_5(10/9)#
#=>log_5((2xx5)/3^2)#
#=>log_5(2xx5)-log_5 3^2# .......(Division Rule)
#=>log_5 2+log_5 5-log_5 3^2# .......(Multiplication Rule)
#=>log_5 2+log_5 5-2log_5 3# .......(Power Rule)

Inserting the given values and remembering that #log_b(b)# is always#=1# we get

#log_5(10/9)=0.4307+1-2xx0.6826#
#=>log_5(10/9)=0.0655#

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