Determine The Value Of The Variable In The Simplified Expressions.1. What Is The Value Of $x$? $\[ \begin{array}{l} G^x + H^{-3} = \frac{1}{g^6} + \frac{1}{h^3} \\ X = \square \end{array} \\]2. What Is The Value Of

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Introduction

In mathematics, solving for the value of a variable in simplified expressions is a crucial skill that helps us understand and manipulate algebraic equations. In this article, we will explore how to determine the value of a variable in a simplified expression using a step-by-step approach. We will use two examples to illustrate this concept.

Example 1: Solving for x

Problem Statement

What is the value of xx?

{ \begin{array}{l} g^x + h^{-3} = \frac{1}{g^6} + \frac{1}{h^3} \\ x = \square \end{array} \}

Step 1: Simplify the Right-Hand Side of the Equation

To solve for xx, we need to simplify the right-hand side of the equation. We can start by finding a common denominator for the two fractions.

1g6+1h3=h3+g6g6h3{ \frac{1}{g^6} + \frac{1}{h^3} = \frac{h^3 + g^6}{g^6h^3} }

Step 2: Rewrite the Equation with the Simplified Right-Hand Side

Now that we have simplified the right-hand side of the equation, we can rewrite the equation as follows:

gx+hβˆ’3=h3+g6g6h3{ g^x + h^{-3} = \frac{h^3 + g^6}{g^6h^3} }

Step 3: Multiply Both Sides of the Equation by g6h3g^6h^3

To eliminate the fraction, we can multiply both sides of the equation by g6h3g^6h^3.

g6h3(gx+hβˆ’3)=g6h3β‹…h3+g6g6h3{ g^6h^3(g^x + h^{-3}) = g^6h^3 \cdot \frac{h^3 + g^6}{g^6h^3} }

Step 4: Simplify the Left-Hand Side of the Equation

Now that we have multiplied both sides of the equation by g6h3g^6h^3, we can simplify the left-hand side of the equation.

gx+6h3+g6hβˆ’3=h3+g6{ g^{x+6}h^3 + g^6h^{-3} = h^3 + g^6 }

Step 5: Subtract h3h^3 from Both Sides of the Equation

To isolate the term with xx, we can subtract h3h^3 from both sides of the equation.

gx+6h3βˆ’h3+g6hβˆ’3=g6{ g^{x+6}h^3 - h^3 + g^6h^{-3} = g^6 }

Step 6: Factor Out h3h^3 from the First Term

Now that we have subtracted h3h^3 from both sides of the equation, we can factor out h3h^3 from the first term.

h3(gx+6βˆ’1)+g6hβˆ’3=g6{ h^3(g^{x+6} - 1) + g^6h^{-3} = g^6 }

Step 7: Subtract g6hβˆ’3g^6h^{-3} from Both Sides of the Equation

To isolate the term with xx, we can subtract g6hβˆ’3g^6h^{-3} from both sides of the equation.

h3(gx+6βˆ’1)=g6βˆ’g6hβˆ’3{ h^3(g^{x+6} - 1) = g^6 - g^6h^{-3} }

Step 8: Divide Both Sides of the Equation by h3h^3

To solve for xx, we can divide both sides of the equation by h3h^3.

gx+6βˆ’1=g6βˆ’g6hβˆ’3h3{ g^{x+6} - 1 = \frac{g^6 - g^6h^{-3}}{h^3} }

Step 9: Simplify the Right-Hand Side of the Equation

Now that we have divided both sides of the equation by h3h^3, we can simplify the right-hand side of the equation.

gx+6βˆ’1=g6(1βˆ’hβˆ’3)h3{ g^{x+6} - 1 = \frac{g^6(1 - h^{-3})}{h^3} }

Step 10: Add 1 to Both Sides of the Equation

To isolate the term with xx, we can add 1 to both sides of the equation.

gx+6=1+g6(1βˆ’hβˆ’3)h3{ g^{x+6} = 1 + \frac{g^6(1 - h^{-3})}{h^3} }

Step 11: Take the Logarithm of Both Sides of the Equation

To solve for xx, we can take the logarithm of both sides of the equation.

log⁑(gx+6)=log⁑(1+g6(1βˆ’hβˆ’3)h3){ \log(g^{x+6}) = \log\left(1 + \frac{g^6(1 - h^{-3})}{h^3}\right) }

Step 12: Use the Power Rule of Logarithms

Now that we have taken the logarithm of both sides of the equation, we can use the power rule of logarithms to simplify the left-hand side of the equation.

(x+6)log⁑(g)=log⁑(1+g6(1βˆ’hβˆ’3)h3){ (x+6)\log(g) = \log\left(1 + \frac{g^6(1 - h^{-3})}{h^3}\right) }

Step 13: Divide Both Sides of the Equation by log⁑(g)\log(g)

To solve for xx, we can divide both sides of the equation by log⁑(g)\log(g).

x+6=log⁑(1+g6(1βˆ’hβˆ’3)h3)log⁑(g){ x+6 = \frac{\log\left(1 + \frac{g^6(1 - h^{-3})}{h^3}\right)}{\log(g)} }

Step 14: Subtract 6 from Both Sides of the Equation

To isolate the value of xx, we can subtract 6 from both sides of the equation.

x=log⁑(1+g6(1βˆ’hβˆ’3)h3)log⁑(g)βˆ’6{ x = \frac{\log\left(1 + \frac{g^6(1 - h^{-3})}{h^3}\right)}{\log(g)} - 6 }

Example 2: Solving for y

Problem Statement

What is the value of yy?

{ \begin{array}{l} \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2} + \frac{1}{d^2} \\ y = \square \end{array} \}

Step 1: Simplify the Right-Hand Side of the Equation

To solve for yy, we need to simplify the right-hand side of the equation. We can start by finding a common denominator for the two fractions.

1c2+1d2=d2+c2c2d2{ \frac{1}{c^2} + \frac{1}{d^2} = \frac{d^2 + c^2}{c^2d^2} }

Step 2: Rewrite the Equation with the Simplified Right-Hand Side

Now that we have simplified the right-hand side of the equation, we can rewrite the equation as follows:

1a2+1b2=d2+c2c2d2{ \frac{1}{a^2} + \frac{1}{b^2} = \frac{d^2 + c^2}{c^2d^2} }

Step 3: Multiply Both Sides of the Equation by c2d2c^2d^2

To eliminate the fraction, we can multiply both sides of the equation by c2d2c^2d^2.

c2d2(1a2+1b2)=c2d2β‹…d2+c2c2d2{ c^2d^2\left(\frac{1}{a^2} + \frac{1}{b^2}\right) = c^2d^2 \cdot \frac{d^2 + c^2}{c^2d^2} }

Step 4: Simplify the Left-Hand Side of the Equation

Now that we have multiplied both sides of the equation by c2d2c^2d^2, we can simplify the left-hand side of the equation.

c2d2(1a2+1b2)=d2+c2{ c^2d^2\left(\frac{1}{a^2} + \frac{1}{b^2}\right) = d^2 + c^2 }

Step 5: Subtract d2d^2 from Both Sides of the Equation

To isolate the term with yy, we can subtract d2d^2 from both sides of the equation.

c2d2(1a2+1b2)βˆ’d2=c2{ c^2d^2\left(\frac{1}{a^2} + \frac{1}{b^2}\right) - d^2 = c^2 }

Step 6: Factor Out d2d^2 from the First Term

Now that we have subtracted d2d^2 from both sides of the equation, we can factor out d2d^2 from the first term.

d2(c2(1a2+1b2)βˆ’1)=c2{ d^2\left(c^2\left(\frac{1}{a^2} + \frac{1}{b^2}\right) - 1\right) = c^2 }

Step 7: Divide Both Sides of the Equation by d2d^2

To solve for yy, we can divide both sides of the equation by d2d^2.

c2(1a2+1b2)βˆ’1=c2d2{ c^2\left(\frac{1}{a^2} + \frac{1}{b^2}\right) - 1 = \frac{c^2}{d^2} }

Step 8: Add 1 to Both Sides of the Equation

To isolate the term with yy, we can add 1 to both sides of the equation.

$[ c2\left(\frac{1}{a2} + \frac{1}{b^2}\right) =