Question

$\displaystyle\sf 2^x - 3^x = \sqrt{6^x - 9^x}$

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Answer 1

Answer:

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Answer 2

Required Answer:-

Given:

$\sf \mapsto {2}^{x} - {3}^{x} = \sqrt{ {6}^{x} - {9}^{x} }$

To find:

.

• The values of x.

Solution:

We have,

$\sf \implies {2}^{x} - {3}^{x} = \sqrt{ {6}^{x} - {9}^{x} }$

Squaring both sides, we get,

$\sf \implies ({2}^{x} - {3}^{x} )^{2} = {6}^{x} - {9}^{x}$

$\sf \implies ({2}^{x} - {3}^{x} )^{2} = {3}^{x} \bigg( \dfrac{ {6}^{x} }{ {3}^{x} } -\dfrac{ {9}^{x} }{ {3}^{x} } \bigg)$

$\sf \implies ({2}^{x} - {3}^{x} )^{2} = {3}^{x}( {2}^{x} - {3}^{x} )$

$\sf \implies{3}^{x}( {2}^{x} - {3}^{x} ) - {( {2}^{x} - {3}^{x} ) }^{2} = 0$

Taking 2ˣ- 3ˣ as common, we get,

$\sf \implies( {2}^{x} - {3}^{x} ) \{ {3}^{x} - ( {2}^{x} - {3}^{x} ) \} = 0$

$\sf \implies( {2}^{x} - {3}^{x} ) (2 \times {3}^{x} - {2}^{x}) = 0$

By zero product rule,

$\sf \implies( {2}^{x} - {3}^{x} ) = 0 \: or \: (2 \times {3}^{x} - {2}^{x}) = 0$

So,

$\sf \implies( {2}^{x} - {3}^{x} ) = 0$

$\sf \implies{2}^{x} = {3}^{x}$

This can only be possible when x = 0 i.e, 2⁰ = 1 and 3⁰ = 1 which is same.

➡ So, x = 0 is a solution!!!

Again,

$\sf \implies 2 \times {3}^{x} - {2}^{x} = 0$

$\sf \implies 2 \times {3}^{x} = {2}^{x}$

$\sf \implies \bigg(\dfrac{2}{3} \bigg)^{x} = 2$

Taking log on both sides, we get,

$\sf \implies \log \bigg(\dfrac{2}{3} \bigg)^{x} = \log(2)$

$\sf \implies x\log \bigg(\dfrac{2}{3} \bigg) = \log(2)$

$\sf \implies x = \dfrac{ \log(2)}{\log \bigg(\dfrac{2}{3} \bigg) }$

$\sf \implies x = \log_{ \frac{2}{3} }(2)$

This is another solution of the question!!!

(Hence Solved)

Answer:

• $\sf x = 0, \log_{ \frac{2}{3} }(2)$