Question

Find the value of k:
$( \frac{2}{3}) ^{3} \times ( \frac{2}{3} ) ^{6} = ( \frac{4}{9} ) ^{2k - 3}$
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Answer 1

Question

Find the value of k:

• ${ \sf\left ( \dfrac{2}{3}\right ) ^{3} \times \left ( \dfrac{2}{3}\right ) ^{6} =\left ( \dfrac{4}{9}\right ) ^{2k - 3} }$

Step by step Explanation :

${ \: \: \: \: : \implies \sf\left ( \dfrac{2}{3}\right ) ^{3} \times \left ( \dfrac{2}{3}\right ) ^{6} =\left ( \dfrac{4}{9}\right ) ^{2k - 3} }$

${ \: \: \: \: : \implies \sf\left ( \dfrac{2}{3}\right ) ^{3} \times \left ( \dfrac{2}{3}\right ) ^{6} =\left ( \dfrac{ {2}^{2} }{ {3}^{2} }\right ) ^{2k - 3} }$

${ \: \: \: \: : \implies \sf\left ( \dfrac{2}{3}\right ) ^{3} \times \left ( \dfrac{2}{3}\right ) ^{6} =\left ( \dfrac{ {2}^{} }{ {3}^{} }\right ) ^{2(2k - 3)} }$

• Bases are same power Should be Equal.

${ \: \: \: \: : \implies \sf \: 3 + 6 = 2(2k - 3) }$

${ \: \: \: \: : \implies \sf \: 3 + 6= 4k - 6}$

${ \: \: \: \: : \implies \sf \: 4k = 6 + 3 + 6}$

${ \: \: \: \: : \implies \sf \: 4k = 15}$

${ \: \: \: \: : \implies \bf \: k = \dfrac{15}{4} }$

Verification

${ \: \: \: \: \mapsto \tiny \sf\left ( \dfrac{2}{3}\right ) ^{3} \times \left ( \dfrac{2}{3}\right ) ^{6} =\left ( \dfrac{4}{9}\right ) ^{2k - 3} }$

• Put k = 15/4

${ \: \: \: \: \mapsto \tiny \sf\left ( \dfrac{8}{27}\right ) \times \left ( \dfrac{64}{729}\right )=\left ( \dfrac{4}{9}\right ) ^{(2 \times \frac{15}{4} )- 3} }$

${ \: \: \: \: \mapsto \tiny \sf \dfrac{512}{19683}=\left ( \dfrac{4}{9}\right ) ^{ \frac{19}{2} } }$

${ \: \: \: \: \mapsto \tiny \sf \dfrac{512}{19683}= \sqrt{ \left ( \dfrac{4}{9}\right )} ^{9} }$

${ \: \: \: \: \mapsto \tiny \sf \dfrac{512}{19683}=\left ( \dfrac{2}{3}\right ) ^{9} }$

${ \: \: \: \: \mapsto \tiny \sf \dfrac{512}{19683}=\dfrac{512}{19683}}$