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Question Number 33004 by Rio Mike last updated on 09/Apr/18

how do i solve for x a) 2^(3−x) +2^x = 6 b) (log_3 x)^2 − 6(log_3 x) + 9=0

$${how}\:{do}\:{i}\:{solve}\: \\ $$$$\:\:\:\:{for}\:{x}\: \\ $$$$\left.{a}\right)\:\mathrm{2}^{\mathrm{3}−{x}} +\mathrm{2}^{{x}} =\:\mathrm{6} \\ $$$$\left.{b}\right)\:\left({log}_{\mathrm{3}} {x}\right)^{\mathrm{2}} \:−\:\mathrm{6}\left({log}_{\mathrm{3}} {x}\right)\:+\:\mathrm{9}=\mathrm{0} \\ $$

Commented by gunawan last updated on 09/Apr/18

a. 2^3 .2^(−x) +2^x =6 let 2^x =a 8a^(−1) +a=6 (8/a)+a=6 a^2 −6a+8=0 (a−2)(a−4)=0 a=2 → 2^x =2 x=1 a=4 2^x =2^2 →x=2 x=(1,2)

$$\mathrm{a}.\:\mathrm{2}^{\mathrm{3}} .\mathrm{2}^{−{x}} +\mathrm{2}^{{x}} =\mathrm{6} \\ $$$$\mathrm{let}\:\mathrm{2}^{{x}} ={a}\: \\ $$$$\mathrm{8}{a}^{−\mathrm{1}} +{a}=\mathrm{6} \\ $$$$\frac{\mathrm{8}}{{a}}+{a}=\mathrm{6} \\ $$$${a}^{\mathrm{2}} −\mathrm{6}{a}+\mathrm{8}=\mathrm{0} \\ $$$$\left({a}−\mathrm{2}\right)\left({a}−\mathrm{4}\right)=\mathrm{0} \\ $$$${a}=\mathrm{2}\:\rightarrow\:\mathrm{2}^{{x}} =\mathrm{2} \\ $$$${x}=\mathrm{1} \\ $$$${a}=\mathrm{4} \\ $$$$\mathrm{2}^{{x}} =\mathrm{2}^{\mathrm{2}} \rightarrow{x}=\mathrm{2} \\ $$$${x}=\left(\mathrm{1},\mathrm{2}\right) \\ $$

Commented by gunawan last updated on 09/Apr/18

b. let log_3 x=a a^2 −6a+9=0 (a−3)(a−3)=0 a=3 log_3 x=3 x=27

$$\mathrm{b}.\:\mathrm{let}\:\mathrm{log}_{\mathrm{3}} {x}={a} \\ $$$${a}^{\mathrm{2}} −\mathrm{6}{a}+\mathrm{9}=\mathrm{0} \\ $$$$\left({a}−\mathrm{3}\right)\left({a}−\mathrm{3}\right)=\mathrm{0} \\ $$$${a}=\mathrm{3} \\ $$$${log}_{\mathrm{3}} {x}=\mathrm{3} \\ $$$${x}=\mathrm{27} \\ $$

Commented by abdo imad last updated on 09/Apr/18

b) (log_3 x)^2 −6 (log_3 x) +9 =0 ⇔ (log_3 (x) −3)^2 =0⇔ log_3 (x)=3 ⇔ ((lnx)/(ln3)) =3 ⇔ ln(x)=ln(27) ⇔x=27 .

$$\left.{b}\right)\:\left({log}_{\mathrm{3}} {x}\right)^{\mathrm{2}} \:−\mathrm{6}\:\left({log}_{\mathrm{3}} {x}\right)\:+\mathrm{9}\:=\mathrm{0}\:\Leftrightarrow\:\left({log}_{\mathrm{3}} \left({x}\right)\:−\mathrm{3}\right)^{\mathrm{2}} \:=\mathrm{0}\Leftrightarrow \\ $$$${log}_{\mathrm{3}} \left({x}\right)=\mathrm{3}\:\:\Leftrightarrow\:\:\frac{{lnx}}{{ln}\mathrm{3}}\:=\mathrm{3}\:\Leftrightarrow\:{ln}\left({x}\right)={ln}\left(\mathrm{27}\right)\:\Leftrightarrow{x}=\mathrm{27}\:. \\ $$

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