Powers of a Fraction: Variable in the exponent of the power

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

We begin with the expression: $\frac{1}{a^{-x}}$.
Our goal is to simplify this expression while converting any negative exponents into positive ones.

• Recall the rule for negative exponents: $a^{-n} = \frac{1}{a^n}$.
• Correspondingly, $\frac{1}{a^{-n}} = a^n$.
• Thus, in our expression $\frac{1}{a^{-x}}$, the negative exponent can be converted and flipped to the numerator by the rule: $\frac{1}{a^{-x}} = a^x$.