Помогитн с математикой Решение показательных уровней

VIII
49^x - 6*7^x - 7 = 0
49^x = (7^2)^x = 7^(2x) = (7^x)^2 = t^2
t^2 - 6*t - 7 = 0
t = -1 и +7
7^x = -1 невозможно;
7^x = +7 при x = 1.

VII
25^x - 6*5^x + 5 = 0
25^x = (5^2)^x = 5^(2x) = (5^x)^2 = t^2
t^2 - 6*t + 5 = 0
t = +1 и +5
7^x = 1 при x = 0;
7^x = 5 при x = log7(5) = ln(5) : ln(7).

VI
4^x - 9*2^x + 8 = 0
4^x = (2^2)^x = 2^(2x) = (2^x)^2 = t^2
t^2 - 9*t + 8 = 0
t = 1 и 8
2^x = 1 при x = 0;
2^x = 8 при x = 3.

V
9^x - 4*3^x + 3 = 0
9^x = (3^2)^x = 3^(2x) = (3^x)^2 = t^2
t^2 - 4*t + 3 = 0
t = 1 и 3
3^x = 1 при x = 0;
3^x = 3 при x = 1.

IV
16^x - 17*4^x + 16 = 0
16^x = (4^2)^x = 4^(2x) = (4^x)^2 = t^2
t^2 - 17*t + 16 = 0
t = 1 и 16
4^x = 1 при x = 0;
4^x = 16 при x = 2.

III
3^(2x^2-x+1) - 2*3^(2x^2-x-1) = 63
(3^2 - 2) * 3^(2x^2-x-1) = 63
3^(2x^2-x-1) = 63 : 7 = 9
2x^2-x-1 = 2
2x^2 - x - 3 = 0
D = 1+4*2*3 = 25
x = (1+-5)/4 = -1 и +3/2.

II
5^(1-2x) + 4*5^(2-2x) - 5^(3-2x) = -0,8
5^(1-2x) + 4*5^1*5^(1-2x) - 5^2*5^(1-2x) = -0,8
(1 + 20 - 25) * 5^(1-2x) = -0,8
5^(1-2x) = (-0,8) : (-4) = 0,8 : 4 = 0,2 = 1/5 = 5^(-1)
1-2x = -1, потому x = 1.

I
2^(12x-1) - 4^(6x-1) + 8^(4x-1) - 16^(3x-1) = 640 есть
2^(12x-1) - 2^(12x-2) + 2^(12x-3) - 2^(12x-4) = 640;
чтобы неизвестным стало 2^(12x-4), вынесем множители:
2^(12x-4)*2^3 - 2^(12x-4)*2^2 + 2^(12x-4)*2^1 - 2^(12x-4) = 640
(8 - 4 + 2 - 1) * 2^(12x-4) = 640
2^(12x-4) = 640 : 5 = 128 = 2^7
12x-4 = 7, потому 12x = 7+4 = 11;
x = 11/12.

1
2^(12x-1)-2^(12x-2)+2^(12x-3)-2^(12x-4)=640
2^(12x)(1/2-1/4+1/8-1/16)=640
2^(12x)*5/16=640
2^(12x)=2048
2^(12x)=2^(11)
12x=11
x=11/12

2
5^(-2x)*(5+100-125)=-0,8
5^(-2x)*(-20)=-0,8
5^(-2x)=4/100
5^(-2x)=5^(-2)
-2x=-2
x=1

4
t>0
4^x=t
t^2-17t+16=0
t1=1
t2=16
4^x=1
x1=0
4^x=16
x2=2

5
t>0
3^x=t
t^2-4t+3=0
t1=1
t2=3
3^x=1
x1=0
3^x=3
x2=1

6
t>0
2^x=t
t^2-9t+8=0
t1=1
t2=8
2^x=1
x1=0
2^x=8
x2=3