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Question Number 172028 by Mikenice last updated on 23/Jun/22

solve: ((log_2 (9−2^(x)) )/(log_2 2^((3−x)) ))=log_2 2

$${solve}: \\ $$$$\frac{{log}_{\mathrm{2}} \left(\mathrm{9}−\mathrm{2}^{\left.{x}\right)} \right.}{{log}_{\mathrm{2}} \mathrm{2}^{\left(\mathrm{3}−{x}\right)} }={log}_{\mathrm{2}} \mathrm{2} \\ $$

Answered by Rasheed.Sindhi last updated on 23/Jun/22

((log_2 (9−2^(x)) )/(log_2 2^((3−x)) ))=log_2 2 ((log_2 (9−2^(x)) )/(log_2 2^((3−x)) ))=1 log_2 (9−2^x )=log_2 2^((3−x)) 9−2^x =2^(3−x) 2^x +2^(3−x) =9 9 is sum of powers of 2 9=8+1 9=2^3 +2^0 2^x +2^(3−x) =2^3 +2^0 2^x +2^(3−x) =2^3 +2^(3−3) x=3

$$\frac{{log}_{\mathrm{2}} \left(\mathrm{9}−\mathrm{2}^{\left.{x}\right)} \right.}{{log}_{\mathrm{2}} \mathrm{2}^{\left(\mathrm{3}−{x}\right)} }={log}_{\mathrm{2}} \mathrm{2} \\ $$$$\frac{{log}_{\mathrm{2}} \left(\mathrm{9}−\mathrm{2}^{\left.{x}\right)} \right.}{{log}_{\mathrm{2}} \mathrm{2}^{\left(\mathrm{3}−{x}\right)} }=\mathrm{1} \\ $$$${log}_{\mathrm{2}} \left(\mathrm{9}−\mathrm{2}^{{x}} \right)={log}_{\mathrm{2}} \mathrm{2}^{\left(\mathrm{3}−{x}\right)} \\ $$$$\mathrm{9}−\mathrm{2}^{{x}} =\mathrm{2}^{\mathrm{3}−{x}} \\ $$$$\mathrm{2}^{{x}} +\mathrm{2}^{\mathrm{3}−{x}} =\mathrm{9} \\ $$$$\mathrm{9}\:{is}\:{sum}\:{of}\:{powers}\:{of}\:\mathrm{2} \\ $$$$\mathrm{9}=\mathrm{8}+\mathrm{1} \\ $$$$\mathrm{9}=\mathrm{2}^{\mathrm{3}} +\mathrm{2}^{\mathrm{0}} \\ $$$$\mathrm{2}^{{x}} +\mathrm{2}^{\mathrm{3}−{x}} =\mathrm{2}^{\mathrm{3}} +\mathrm{2}^{\mathrm{0}} \\ $$$$\mathrm{2}^{{x}} +\mathrm{2}^{\mathrm{3}−{x}} =\mathrm{2}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}−\mathrm{3}} \\ $$$${x}=\mathrm{3} \\ $$

Commented by Mikenice last updated on 23/Jun/22

yhanks sir

$${yhanks}\:{sir} \\ $$

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