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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
howdoisolveforxa)23−x+2x=6b)(log3x)2−6(log3x)+9=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)
a.23.2−x+2x=6let2x=a8a−1+a=68a+a=6a2−6a+8=0(a−2)(a−4)=0a=2→2x=2x=1a=42x=22→x=2x=(1,2)
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
b.letlog3x=aa2−6a+9=0(a−3)(a−3)=0a=3log3x=3x=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 .
b)(log3x)2−6(log3x)+9=0⇔(log3(x)−3)2=0⇔log3(x)=3⇔lnxln3=3⇔ln(x)=ln(27)⇔x=27.

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