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