Powers of a Fraction: System of equations with no solution

To solve the problem, we need to simplify the expression $\left(\frac{11\times9}{10\times12}\right)^{x+a}$ and write it in the form requested in the question.

We begin by using the exponent rule: $(\frac{a}{b})^n = \frac{a^n}{b^n}$. Applying this rule here:

$<span class="katex"> \left(\frac{11\times9}{10\times12}\right)^{x+a} = \frac{(11\times9)^{x+a}}{(10\times12)^{x+a}} </span>$

Next, we can simplify the expression further by applying the power over a product rule: $(ab)^n = a^n \times b^n$.

Applying this rule to both the numerator and denominator gives us:

Numerator: $(11\times9)^{x+a} = 11^{x+a} \times 9^{x+a}$

Denominator: $(10\times12)^{x+a} = 10^{x+a} \times 12^{x+a}$

Therefore, the entire expression becomes:

$<span class="katex"> \frac{11^{x+a} \times 9^{x+a}}{10^{x+a} \times 12^{x+a}} </span>$

This matches the given answer. Thus, the solution to the question is:

$\frac{11^{x+a}\times9^{x+a}}{10^{x+a}\times12^{x+a}}$

Let's start by examining the expression given in the question:

$\left(\frac{13}{7\times6\times3}\right)^{x+y}$

This expression is a power of a fraction. There is a general rule in exponents which states:

$\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$

Using this rule, we will apply it to our original expression.

Given, $a = 13$, $b = 7\times6\times3$, and $n = x+y$, we can rewrite our expression as:

$\frac{13^{x+y}}{(7\times6\times3)^{x+y}}$

The solution to the question is:

$\frac{13^{x+y}}{(7\times6\times3)^{x+y}}$

Let's analyze the expression we are given:

$\left(\frac{2\times4\times6}{7\times8\times9}\right)^{3x}$

The expression is a power of a fraction. The rule for powers of a fraction is that each component of the fraction must be raised to the power separately. This can be expressed as:

$\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$

Applying this rule to our expression, we have:

• The numerator inside the power: $2 \times 4 \times 6$
• The denominator inside the power: $7 \times 8 \times 9$

Therefore, raising each part to the power $3x$ gives us:

$\frac{(2\times4\times6)^{3x}}{(7\times8\times9)^{3x}}$

Thus, the simplified expression for the given equation is:

$\frac{\left(2\times4\times6\right)^{3x}}{\left(7\times8\times9\right)^{3x}}$

The solution to the question is: $\frac{\left(2\times4\times6\right)^{3x}}{\left(7\times8\times9\right)^{3x}}$

The given expression is:
$\left(\frac{7\times11\times19}{3\times12\times15}\right)^{-4}$

To solve this expression, we need to apply the rules of exponents, specifically the rule for powers of a fraction. For any fraction$\left(\frac{a}{b}\right)^{-n}$, the expression is equivalent to$\left(\frac{b}{a}\right)^n$.
Therefore, negative exponents indicate that the fraction should be flipped and raised to the positive of that exponent.

Substitute the terms into this formula:
1. Flip the fraction: $\left(\frac{3\times12\times15}{7\times11\times19}\right)$
2. Raise both numerator and denominator to the power of 4:
Thus, we have:
$\left(\frac{3\times12\times15}{7\times11\times19}\right)^{4}$

Now evaluating each term individually:
- In the numerator:
- $3^4\times12^4\times15^4$
- In the denominator:
- $7^4\times11^4\times19^4$

Applying the negative exponent rule, each individual factor in both numerator and denominator should be inverted, altering the exponents to negative:
1. Numerator becomes: $3^{-4}\times12^{-4}\times15^{-4}$
2. Denominator becomes: $7^{-4}\times11^{-4}\times19^{-4}$

Rewriting the expression, we achieve:
$\frac{7^{-4}\times11^{-4}\times19^{-4}}{3^{-4}\times12^{-4}\times15^{-4}}$

This matches precisely the provided solution.

The solution to the question is:$\frac{7^{-4}\times11^{-4}\times19^{-4}}{3^{-4}\times12^{-4}\times15^{-4}}$