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Question Number 170032 by Mastermind last updated on 14/May/22

((8^x −2^x )/(6^x −3^x ))=2 , find x ? Mastermind

$$\frac{\mathrm{8}^{{x}} −\mathrm{2}^{{x}} }{\mathrm{6}^{{x}} −\mathrm{3}^{{x}} }=\mathrm{2}\:,\:{find}\:{x}\:? \\ $$$$ \\ $$$${Mastermind} \\ $$

Commented by Mastermind last updated on 14/May/22

your solution please

$${your}\:{solution}\:{please} \\ $$

Commented by Rasheed.Sindhi last updated on 14/May/22

x=1

$${x}=\mathrm{1} \\ $$

Answered by Rasheed.Sindhi last updated on 14/May/22

((8^x −2^x )/(6^x −3^x ))=2 , find x ? ((2^x (4^x −1))/(3^x (2^x −1)))=2 ((2^x (2^(2x) −1))/(3^x (2^x −1)))=2 ((2^x (2^x −1)(2^x +1))/(3^x (2^x −1)))=2 ;2^x ≠1⇒x≠0^★ ((2^x (2^x +1))/3^x )=2 Assuming x a whole number 2^x ∙((2^x +1)/3^x )=2=1×2 (2^x =1 ∧ ((2^x +1)/3^x )=2)∨(2^x =2 ∧ ((2^x +1)/3^x )=1) (2^x =2^0 ∧ ((2^0 +1)/3^0 )=2)∨(2^x =2^1 ∧ ((2^1 +1)/3^1 )=1) x≠0^★ Hence x=1 (Only one solution in whole numbers)

$$\frac{\mathrm{8}^{{x}} −\mathrm{2}^{{x}} }{\mathrm{6}^{{x}} −\mathrm{3}^{{x}} }=\mathrm{2}\:,\:{find}\:{x}\:? \\ $$$$\frac{\mathrm{2}^{{x}} \left(\mathrm{4}^{{x}} −\mathrm{1}\right)}{\mathrm{3}^{{x}} \left(\mathrm{2}^{{x}} −\mathrm{1}\right)}=\mathrm{2} \\ $$$$\frac{\mathrm{2}^{{x}} \left(\mathrm{2}^{\mathrm{2}{x}} −\mathrm{1}\right)}{\mathrm{3}^{{x}} \left(\mathrm{2}^{{x}} −\mathrm{1}\right)}=\mathrm{2} \\ $$$$\frac{\mathrm{2}^{{x}} \cancel{\left(\mathrm{2}^{{x}} −\mathrm{1}\right)}\left(\mathrm{2}^{{x}} +\mathrm{1}\right)}{\mathrm{3}^{{x}} \cancel{\left(\mathrm{2}^{{x}} −\mathrm{1}\right)}}=\mathrm{2}\:;\mathrm{2}^{{x}} \neq\mathrm{1}\Rightarrow{x}\neq\mathrm{0}^{\bigstar} \\ $$$$\frac{\mathrm{2}^{{x}} \left(\mathrm{2}^{{x}} +\mathrm{1}\right)}{\mathrm{3}^{{x}} }=\mathrm{2} \\ $$$${Assuming}\:{x}\:{a}\:{whole}\:{number} \\ $$$$\mathrm{2}^{{x}} \centerdot\frac{\mathrm{2}^{{x}} +\mathrm{1}}{\mathrm{3}^{{x}} }=\mathrm{2}=\mathrm{1}×\mathrm{2} \\ $$$$\left(\mathrm{2}^{{x}} =\mathrm{1}\:\wedge\:\:\frac{\mathrm{2}^{{x}} +\mathrm{1}}{\mathrm{3}^{{x}} }=\mathrm{2}\right)\vee\left(\mathrm{2}^{{x}} =\mathrm{2}\:\wedge\:\frac{\mathrm{2}^{{x}} +\mathrm{1}}{\mathrm{3}^{{x}} }=\mathrm{1}\right) \\ $$$$\left(\mathrm{2}^{{x}} =\mathrm{2}^{\mathrm{0}} \:\wedge\:\:\frac{\mathrm{2}^{\mathrm{0}} +\mathrm{1}}{\mathrm{3}^{\mathrm{0}} }=\mathrm{2}\right)\vee\left(\mathrm{2}^{{x}} =\mathrm{2}^{\mathrm{1}} \:\wedge\:\frac{\mathrm{2}^{\mathrm{1}} +\mathrm{1}}{\mathrm{3}^{\mathrm{1}} }=\mathrm{1}\right) \\ $$$${x}\neq\mathrm{0}^{\bigstar} \: \\ $$$${Hence}\:{x}=\mathrm{1}\:\left({Only}\:{one}\:{solution}\:{in}\:{whole}\:{numbers}\right) \\ $$

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