Find the value of ${log}_{2} . {log}_{3} 9$

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Solution

9 can be written as $3^{2}$

So, ${log}_{2} . {log}_{3} . 3^{2}$

= ${log}_{2} . 2 {log}_{3} 3$ {since ${log}_{3} 3 = 1$ }

= ${log}_{2} 2$

= 1

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Similar questions

Find the value of ${log}_{2} . {log}_{3} 9$

l o g_{3} 9 + l o g_{5} 25 + l o g_{2} 8

\frac{{log}_{2} a}{4} = \frac{{log}_{2} b}{6} = \frac{{log}_{2} c}{3 p}

a^{2} b^{2} c = 1

- p = \frac{20}{m}

m

l o g_{2 \sqrt{3}} 1728

{log}_{2} 32

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