Exponential equations grade 11 math

Please help with grade 11 math homework:

I am supposed to determine the exact value of x of the following:

a. 32x = 5(3x) + 36
b. (1/8)^(x-3) = 2*16^(2x+1)
c. 3^(x²+20) = (1/27)^(3x)

Can you tell me exactly how I can do this for each of the above?

Thank you so much for your time and your help! :)

3^(2x) = 5 * 3^(x) + 36

Let 3^(x) = t

t^2 = 5 * t + 36
t^2 - 5t - 36 = 0
t = (5 +/- sqrt(25 + 144)) / 2
t = (5 +/- 13) / 2
t = -8/2 , 18/2
t = -4 , 9
3^x = -4 , 9
3^x cannot be negative for real values of x
3^x = 9
3^x = 3^2
x = 2



(1/8)^(x - 3) = 2 * 16^(2x + 1)
(2^-3)^(x - 3) = 2 * (2^4)^(2x + 1)
2^(-3x + 9) = 2^(1 + 4 * (2x + 1))
9 - 3x = 1 + 8x + 4
9 - 3x = 5 + 8x
9 - 5 = 8x + 3x
4 = 11x
x = 4/11


3^(x^2 + 20) = (1/27)^(3x)
3^(x^2 + 20) = 3^(-3 * 3x)
x^2 + 20 = -9x
x^2 + 9x + 20 = 0
x = (-9 +/- sqrt(9^2 - 4 * 20)) / 2
x = (-9 +/- sqrt(81 - 80)) / 2
x = (-9 +/- 1) / 2
x = -10/2 , -8/2
x = -5 , -4

C)........1 / 27 = 1 / 3^3 = 3 ^ (-3)

3 ^ (x² + 20) = 3 ^ [ -3(3x) ]

3 ^ (x² + 20) = 3 ^ (-9x).............When bases are alike...set exponents equal

x² + 20 = -9x

x² + 9x + 20 = 0

(x + 5)(x + 4) = 0

x + 5 = 0.....x + 4 = 0

x = - 5.......x = - 4................ANSWER

Mathcerely, Robert Jones "Teacher/Tutor of Fine Students"
Moved Florida to France June 2011 but still tutor world-wide (USA , UK ,
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********* A PRESENT FOR YOU
Ask someone to write a number, say a five-digit number.
(Can be a 2-digit.....3-digit.....4-digit....ect..…
Suppose the number written is
57836
Now, without showing the asker, you write a number on