Answer :
Sure, let's solve each of these problems step-by-step.
### Problem 14: Express [tex]\(\frac{27 x^{-3} y^2}{\left(3 x^{-2} y^4\right)^2}\)[/tex] without negative indices
First, let's simplify the expression [tex]\(\frac{27 x^{-3} y^2}{\left(3 x^{-2} y^4\right)^2}\)[/tex].
Define [tex]\( y = 4 - |x| \)[/tex].
Rewriting the expression:
[tex]\[ \frac{27 x^{-3} y^2}{\left(3 x^{-2} y^4\right)^2} \][/tex]
Let's start by simplifying the denominator:
[tex]\[ (3 x^{-2} y^4)^2 = 3^2 \cdot (x^{-2})^2 \cdot (y^4)^2 = 9 \cdot x^{-4} \cdot y^8 \][/tex]
Now, plug this back into the original expression:
[tex]\[ \frac{27 x^{-3} y^2}{9 x^{-4} y^8} \][/tex]
Combine the coefficients and the variables separately:
[tex]\[ \frac{27}{9} \cdot \frac{x^{-3}}{x^{-4}} \cdot \frac{y^2}{y^8} = 3 \cdot x^{1} \cdot y^{-6} \][/tex]
Since [tex]\(y = 4 - |x|\)[/tex], substitute [tex]\(y\)[/tex] back into the expression:
[tex]\[ 3 x \cdot (4 - |x|)^{-6} \][/tex]
So, the simplified answer is:
[tex]\[ \boxed{\frac{3x}{(4 - |x|)^6}} \][/tex]
### Problem 15: Solve the equation [tex]\(2^{3 x^2} \times 2^{4 x} = \frac{2^2}{2^{-2}}\)[/tex] for [tex]\(x\)[/tex]
First, simplify the right-hand side of the equation:
[tex]\[ \frac{2^2}{2^{-2}} \][/tex]
Recall that [tex]\(\frac{a^m}{a^n} = a^{m-n}\)[/tex]:
[tex]\[ \frac{2^2}{2^{-2}} = 2^{2 - (-2)} = 2^{2 + 2} = 2^4 \][/tex]
The equation is now:
[tex]\[ 2^{3 x^2} \times 2^{4 x} = 2^4 \][/tex]
Using the property [tex]\(a^m \times a^n = a^{m+n}\)[/tex], we can combine the exponents on the left-hand side:
[tex]\[ 2^{3 x^2 + 4 x} = 2^4 \][/tex]
Since the bases are the same, the exponents must be equal:
[tex]\[ 3 x^2 + 4 x = 4 \][/tex]
This is a quadratic equation; let's rewrite it in standard form:
[tex]\[ 3 x^2 + 4 x - 4 = 0 \][/tex]
We can solve this quadratic equation using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our equation [tex]\(3x^2 + 4x - 4 = 0\)[/tex]:
[tex]\[ a = 3, \quad b = 4, \quad c = -4 \][/tex]
Plug these values into the quadratic formula:
[tex]\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{16 + 48}}{6} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{64}}{6} \][/tex]
[tex]\[ x = \frac{-4 \pm 8}{6} \][/tex]
This gives us two solutions:
1. [tex]\(x = \frac{-4 + 8}{6} = \frac{4}{6} = \frac{2}{3}\)[/tex]
2. [tex]\(x = \frac{-4 - 8}{6} = \frac{-12}{6} = -2\)[/tex]
So, the solutions are:
[tex]\[ \boxed{-2 \text{ and } \frac{2}{3}} \][/tex]
### Problem 14: Express [tex]\(\frac{27 x^{-3} y^2}{\left(3 x^{-2} y^4\right)^2}\)[/tex] without negative indices
First, let's simplify the expression [tex]\(\frac{27 x^{-3} y^2}{\left(3 x^{-2} y^4\right)^2}\)[/tex].
Define [tex]\( y = 4 - |x| \)[/tex].
Rewriting the expression:
[tex]\[ \frac{27 x^{-3} y^2}{\left(3 x^{-2} y^4\right)^2} \][/tex]
Let's start by simplifying the denominator:
[tex]\[ (3 x^{-2} y^4)^2 = 3^2 \cdot (x^{-2})^2 \cdot (y^4)^2 = 9 \cdot x^{-4} \cdot y^8 \][/tex]
Now, plug this back into the original expression:
[tex]\[ \frac{27 x^{-3} y^2}{9 x^{-4} y^8} \][/tex]
Combine the coefficients and the variables separately:
[tex]\[ \frac{27}{9} \cdot \frac{x^{-3}}{x^{-4}} \cdot \frac{y^2}{y^8} = 3 \cdot x^{1} \cdot y^{-6} \][/tex]
Since [tex]\(y = 4 - |x|\)[/tex], substitute [tex]\(y\)[/tex] back into the expression:
[tex]\[ 3 x \cdot (4 - |x|)^{-6} \][/tex]
So, the simplified answer is:
[tex]\[ \boxed{\frac{3x}{(4 - |x|)^6}} \][/tex]
### Problem 15: Solve the equation [tex]\(2^{3 x^2} \times 2^{4 x} = \frac{2^2}{2^{-2}}\)[/tex] for [tex]\(x\)[/tex]
First, simplify the right-hand side of the equation:
[tex]\[ \frac{2^2}{2^{-2}} \][/tex]
Recall that [tex]\(\frac{a^m}{a^n} = a^{m-n}\)[/tex]:
[tex]\[ \frac{2^2}{2^{-2}} = 2^{2 - (-2)} = 2^{2 + 2} = 2^4 \][/tex]
The equation is now:
[tex]\[ 2^{3 x^2} \times 2^{4 x} = 2^4 \][/tex]
Using the property [tex]\(a^m \times a^n = a^{m+n}\)[/tex], we can combine the exponents on the left-hand side:
[tex]\[ 2^{3 x^2 + 4 x} = 2^4 \][/tex]
Since the bases are the same, the exponents must be equal:
[tex]\[ 3 x^2 + 4 x = 4 \][/tex]
This is a quadratic equation; let's rewrite it in standard form:
[tex]\[ 3 x^2 + 4 x - 4 = 0 \][/tex]
We can solve this quadratic equation using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our equation [tex]\(3x^2 + 4x - 4 = 0\)[/tex]:
[tex]\[ a = 3, \quad b = 4, \quad c = -4 \][/tex]
Plug these values into the quadratic formula:
[tex]\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{16 + 48}}{6} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{64}}{6} \][/tex]
[tex]\[ x = \frac{-4 \pm 8}{6} \][/tex]
This gives us two solutions:
1. [tex]\(x = \frac{-4 + 8}{6} = \frac{4}{6} = \frac{2}{3}\)[/tex]
2. [tex]\(x = \frac{-4 - 8}{6} = \frac{-12}{6} = -2\)[/tex]
So, the solutions are:
[tex]\[ \boxed{-2 \text{ and } \frac{2}{3}} \][/tex]
