Please help me for math ill give branliest

(5^3 * 2^-6)^-2 = 2^w/c^x

we can first reciprocate both sides, note that z^-b = 1/z^b where z is real and cannot equal zero and b is real. So reciprocating just turns it into z^b, in this case, imagine z = (5^3 * 2^-6) and b = 2.
Note that reciprocal also means ‘to the power of -1’ and using exponent rules means (z^-b)^-1 = z^(-1*b*-1) = z^b

Therefore,

[(5^3 * 2^-6)^-2]^-1 = [2^w/c^x]^-1

z^b = c^x/2^w

(5^3 * 2^-6)^2 = c^x/2^w

using exponent rules, including reciprocation,

5^6*2^-12 = c^x/2^w

and again,

5^6/2^12 = c^x/2^w

now 5 and 2 are relatively prime (do not have any common factors); in fact, they are both distinct prime numbers, so they would not share any factors. This also means that the numerator is in its ‘prime factorised’ form, as well as the denominator. I am assuming we are looking for integer solutions.

Therefore, the solutions naturally come out.

c = 5

x = 6

w = 12

Multiplication property of zero

Num of plants = (original#ofplants)*(1.6)^t