HOW TO EVALUATE SIMPLE LOGARITHMS PROBLEMS

HOW TO EVALUATE LOGARITHMS PROBLEMS

The following exercises have been extracted from the New General Mathematics Book 3 for Senior Secondary Schools. In my last tutorial on logarithm, i discussed on the three major laws of logarithms and also gave some examples. click here to be redirected back.

Photo Showing the 3 Laws of Logarithm, this laws will be used to

 evaluate the exercises below.

Express the Following as Logarithms of Single Numbers.

(a)  $\log{5}+\log{6}$, using the product law.
   
     $=\log{5\times{6}}=\log{30}$

(b) $\log{8}-\log{6}$, using the quotient law
   
     $\log\frac{8}{6}=\log\frac{4}{3}=\log{1}\frac{1}{3}$


(c)  $3\log{6}$ Using the power law
  
     $=\log{6^3}=\log{216}$

(d)  $\frac{1}{2}\log{49}=\frac{1}{2}\log{7^2}=\log{7}$



(e)  $-2\log{4}=\log\frac{1}{4^2}=\log\frac{1}{16}$

(f)  $1+\log{5}=$ remember $\log{10}=1$, therefore

     $\log{10}+\log{5}=\log{(10\times{5})}=\log{50}$

(g)  $1-\log{2}=\log{10}-\log{2}=\log\frac{10}{2}=\log{5}$

(h)  $\log{18}-2\log{2}=\log{18}-\log{2^2}$

     $=\log{18}-\log{4}=\log\frac{18}{4}=\log\frac{9}{2}=\log{4}\frac{1}{2}$

(i)  $2-2\log{5}=\log{100}-\log{5^2}=\log{100}-\log{25}$

     $=\log\frac{100}{25}=\log{4}$

(j)  $\frac{3}{5}\log{32}=\log{32}^{\frac{3}{5}}=\log{(2^5)}^{\frac{3}{5}}$
 
     $\log{2^3}=\log{8}$.