Solve the equation 6(9)^x-17(3^x)+5=0 and determine all solutions to the equation: 2^(3x)+2^(2x)-26(2^x)+24=0

Question

Solve the equation 

6(9)^x-17(3^x)+5=0 and determine all solutions to the equation: 2^(3x)+2^(2x)-26(2^x)+24=0

Answer 1

6(9^x) - 17(3^x) + 5 = 0

6(3^x)² - 17(3^x) + 5 = 0

Let u = 3^x. Then we have 6u² - 17u + 5 = 0

(3u - 1)(2u - 5) = 0

u = 1/3 or 5/2

3^x = 3^-1 or 3^x = 5/2

x = -1 or log₃(5/2)

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(2^x)³ + (2^x)² - 26(2^x) + 24 = 0

Let u = 2^x. Then u³ + u² - 26u + 24 = 0

By trial and error, u = 1 is a solution. So, u - 1 is a factor.

Dividing out by u - 1, we get (u-1)(u² + 2u - 24) = 0

(u - 1)(u + 6)(u - 4) = 0

u = 1, 4, or -6

2^x = 1, 2^x = 2², or 2^x = -6

2^x can't equal -6 since 2^x is always positive.

So, x = 0 or x = 2

Answer 2

6(9)^x-17(3)^x+ 5 = 0; (3·3^x - 1)(2·3^x - 5) = 0;

3·3^x - 1 = 0; 3^x = 3^-1, x = - 1; 2·3^x - 5 = 0; 3^x = 2.5; xln3 = ln2.5; x = ln2.5/ln3;

Answer: {- 1, ln2.5/ln3}.

2^3x + 2^2x - 26·2^x + 24 = 0; 2^x = u > 0;

u³ + u² - 26u + 24 = 0; (u³ - u) + (u² - u) - 24(u - 1) = 0; (u - 1)(u² + u + u - 24) = 0;

u - 1 = 0 or (u² + 2u - 24) = 0; u = 1 or u = - 1 + √25 = 4 (u = - 1 - √25 = - 6 < 0 we have to regect).

3^x = 1; x = 0; 3^x = 4; x = ln4/ln3