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Question Number 19382 by chernoaguero@gmail.com last updated on 10/Aug/17

If log_2 (9^(x−1) +7)−log_2 (3^(x−1) +1)=2,   then the values of x are

$$\mathrm{If}\:\mathrm{log}_{\mathrm{2}} \left(\mathrm{9}^{{x}−\mathrm{1}} +\mathrm{7}\right)−\mathrm{log}_{\mathrm{2}} \left(\mathrm{3}^{{x}−\mathrm{1}} +\mathrm{1}\right)=\mathrm{2},\: \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:{x}\:\mathrm{are} \\ $$

Answered by khamizan833@yahoo.com last updated on 10/Aug/17

x = (1,2)

$${x}\:=\:\left(\mathrm{1},\mathrm{2}\right) \\ $$

Answered by ajfour last updated on 10/Aug/17

let 3^x =t  ⇒ (((t^2 /9)+7)/((t/3)+1))=4        t^2 +63=12t+36      t^2 −12t+27=0      (t−3)(t−9)=0  ⇒  t=3^x =3, 9  ⇒  x=1 or x=2 .

$$\mathrm{let}\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} =\mathrm{t} \\ $$$$\Rightarrow\:\frac{\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{9}}+\mathrm{7}}{\frac{\mathrm{t}}{\mathrm{3}}+\mathrm{1}}=\mathrm{4}\:\: \\ $$$$\:\:\:\:\mathrm{t}^{\mathrm{2}} +\mathrm{63}=\mathrm{12t}+\mathrm{36} \\ $$$$\:\:\:\:\mathrm{t}^{\mathrm{2}} −\mathrm{12t}+\mathrm{27}=\mathrm{0} \\ $$$$\:\:\:\:\left(\mathrm{t}−\mathrm{3}\right)\left(\mathrm{t}−\mathrm{9}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{t}=\mathrm{3}^{\boldsymbol{\mathrm{x}}} =\mathrm{3},\:\mathrm{9} \\ $$$$\Rightarrow\:\:\boldsymbol{\mathrm{x}}=\mathrm{1}\:\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{x}}=\mathrm{2}\:. \\ $$

Commented by chernoaguero@gmail.com last updated on 10/Aug/17

Thank sir

$${Thank}\:{sir} \\ $$

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