Tìm x: a, 2.3^x-405=3^x-1 b,(1/81)^x .27^2x =(-9)^4

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1. `a)`

   Ta có :

   `2.3^x – 405 = 3^{x-1}`

   `⇒ 2.3^x – 405 = 3^x . 1/3`

   `⇒ (2.3^x) – ( 3^x . 1/3) = 405`

   `⇒ 3^x(2 – 1/3) = 405`

   `⇒3^x . 5/3 = 405`

   `⇒3^x =243`

   `⇒3^x = 3^5`

   `⇒ x = 5`

   `b)`

   `(1/81)^x .  27^{2x} = (-9)^4`

   `⇒  (1/81)^x . 729^x = (-9)^4`

   `⇒ (729/81.)^x = (-9)^4`

   `⇒ 9^x = (-9)^4` 

   `⇒ x = 4`

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2. `a) 2.3^x-405=3^(x-1)`

   `<=> 3^x. 3^(-1)-2.3^x=-405`

   `<=> 3^x(1/3-2)=-405`

   `<=> 3^x. (-5/3)=-405`

   `<=> 3^x=-405.(-3/5)`

   `<=> 3^x=243`

   `<=> 3^x=3^5`

   `<=> x=5`

   Vậy `x=5`

   `b) (1/81)^x. 27^(2x)=(-9)^4`

   `<=> ((1/3)^4)^x. (3^3)^(2x)=9^4`

   `<=> (1/3)^(4x). 3^(6x)=(3^2)^4`

   `<=> 3^(-4x). 3^(6x)=3^8`

   `<=> 3^(2x)=3^8`

   `<=> 2x=8`

   `<=> x=4`

   Vậy `x=4`

    

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