for what values of x does $25^{x}=5^{x^{2}-3}$?\n$x = - 3,x = 1$\n$x=-1,x = 3$\n$x=-\frac{3}{2},x =…

for what values of x does $25^{x}=5^{x^{2}-3}$?\n$x = - 3,x = 1$\n$x=-1,x = 3$\n$x=-\frac{3}{2},x = 2$\n$x=-2,x=\frac{3}{2}$

for what values of x does $25^{x}=5^{x^{2}-3}$?\n$x = - 3,x = 1$\n$x=-1,x = 3$\n$x=-\frac{3}{2},x = 2$\n$x=-2,x=\frac{3}{2}$

Answer

Explanation:

Step1: Rewrite 25 as 5^2

Since (25 = 5^2), the equation (25^{x}=5^{x^{2}-3}) can be rewritten as ((5^{2})^{x}=5^{x^{2}-3}). According to the power - of - a - power rule ((a^{m})^{n}=a^{mn}), we have (5^{2x}=5^{x^{2}-3}).

Step2: Set the exponents equal

If (a^{m}=a^{n}), then (m = n) for (a>0,a\neq1). So, (2x=x^{2}-3).

Step3: Rearrange to a quadratic equation

Rearrange (2x=x^{2}-3) to the standard quadratic form (ax^{2}+bx + c = 0). We get (x^{2}-2x - 3=0).

Step4: Factor the quadratic equation

Factor (x^{2}-2x - 3) as ((x - 3)(x+1)=0).

Step5: Solve for x

Set each factor equal to zero: (x - 3=0) gives (x = 3), and (x + 1=0) gives (x=-1).

Answer:

(x=-1,x = 3)