Question

Me ajudem
resolva as equações exponenciais

1) (3^2x-7)³÷9^x+1=(3^3x-1)⁴
2)2^3x+2÷8^2x-7=4^x-1

Answer 1

Boa noite Pedrojoao

1)  

3^(2x - 7)^3/9^(x + 1) = 3^(3x - 1)^4 

3^(6x - 21)/3^(2x + 2) = 3^(12x - 4)

6x - 21 - 2x - 2 = 12x - 4
4x - 23 = 12x - 4
12x - 4x = -23 + 4
8x = -19
x = -19/8 

2)

2^(3x + 2)/(8^(2x - 7) = 4^(x - 1)

2^(3x + 2)/(2^(6x - 21) = 2^(2x - 2)

3x + 2 - 6x + 21 = 2x - 2
5x = 25
x = 5 

Answer 2

$\frac{(3^{2x-7})^3}{9^{x+1}}=(3^{3x-1})^4\\ \\ \frac{3^{6x-21}}{(3^2)^{x+1}}=3^{12x-4} \\ \\ 6x-21 -[2(x+1)]=12x-4\\ \\ 6x - 21 -[2x+2] = 12x -4\\ \\ 6x - 21-2x-2=12x-4 \\ \\ 4x -23 = 12x - 4 \\ \\ -8x = 19 \\ \\ x = \frac{-19}{8}$

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$\frac{2^{3x+2}}{8^{2x-7}}=4^{x-1} \\ \\ \frac{2^{3x+2}}{(2^3)^{2x-7}}=(2^2)^{x-1} \\ \\\frac{2^{3x+2}}{2^{6x-21}}=2^{2x-2}\\ \\ 3x + 2 -(6x -21) = 2x -2 \\ \\ 3x + 2 -6x + 21 = 2x - 2 \\ \\ -3x + 23 = 2x - 2 \\ \\ -5x = -25 \\ \\ x = 5$