Solve and check 5 = 25^x/2?

Question 1
5 = 5^x
x = 1

Question 2
Reading question as:-
(3/2)^(m/2) = 4/9
(3/2)^m = 16/81
(2/3)^(- m) = (2/3)^4
- m = 4
m = - 4

n^3 - 9n^2 - n + 9 = 0 factor by capacity of grouping (n^3 - 9n^2) - (n - 9) = 0 n^2 (n - 9) - a million(n - 9) = 0 (n^2 - a million)(n - 9) = 0 (n + a million)(n - a million)(n - 9) = 0 n + a million = 0 ==> n = -a million n - a million = 0 ==> n = a million n - 9 = 0 ==> n = 9 so the two n = a million , -a million, or 9 ***** (4x + 9y)^2 = (squaring binomials) 16x^2 + 36xy + 36xy + 81y^2 = 16x^2 + 72xy + 81y^2 ***** -25x^2 - 115x + 210 = -5(5x-7)(x+6) -25x^2 - 115x + 210 = -5(5x^2 +23x - 40 two) -25x^2 - 115x + 210 = -25x^2 - 115x + 210 yep, that is top ***** GCF: -4m^2n^2, leaving n + 9 you skipped over a unfavourable sign -4m^2 n^3 - 36m^2 n^2 = -4m^2 n^2(n + 9) *** skipped over an x^2: 9x(-x + 2) = -9x^2 + 18x

Q1:
5 = 25^ x/2
5 = 5^x because 25 = 5^2
1= x

Q2:
(3/2)^m/2 = 4/9
Times 2 on both side exponents. It makes:
(3/2)^m = 16/81
(3/2)^m = (2^4 / 3^4)
(3/2)^m = 1 / [(3/2)^4]
(3/2)^m = (3/2)^ -4
m = -4

5 = 25^ x/2
5 = 5^x
1= x


(3/2)^m/2 = 4/9
Squaring both sides:
(3/2)^m = 16/81
(3/2)^m = (2/3)^4
(3/2)^m = 1 / [(3/2)^4]
(3/2)^m = (3/2)^-4
m = -4

I'm assuming brackets required.

1. 5 = 25^(x/2)

5 = (5^2)^(x/2)

5^1 = 5^(x)
Therefore, x = 1

2. (3/2)^(m/2) = 4/9
= 2^2/3^2 = (2/3)^2 = 1/(3/2)^2 = (3/2)^(-2)
Therefore, m/2 = -2, so, m = -4.

hmmm let me solve it but i might be wrong but the other guys i think is wrong
5 = 25^x/2
5 = (root) 25^x
5 = 5 ^x
then i dont know the logarithmic thing , i cant remember the formula

5 = 25^x/2
Squaring both sides: 25 = 25^x
x = 1


(3/2)^m/2 = 4/9
Squaring both sides: (3/2)^m = 16/81
(3/2)^m = (2/3)^4
(3/2)^m = 1 / ((3/2)^4)
(3/2)^m = (3/2)^-4
m = -4