Treat this with regular exponent rules.

\(\displaystyle \frac{x^{3.78}}{x^{2.58}}=x^{3.78-2.58}=x^{1.2}\)

\(\displaystyle \frac{15x^2y^5c^3}{5xy^6c^3} \rightarrow 3xy^{-1}\rightarrow 3x*\frac{1}{y}=\boldsymbol{\frac{3x}{y}}\)

When multiplying exponential, the exponents always add. While the 2 in the front of the first exponential might throw you off, you may disregard it initially.

\(\displaystyle (2)(x^{7.2}x^{-6.2})\)

Which simplifies to

\(\displaystyle (2)x^{7.2-6.2}=(2)x^1\)

Our final answer is \(\displaystyle 2x\)

First, we need to simplify the numerator. First term, \(\displaystyle (x^2)^5\) can be simplified to \(\displaystyle x^{10}\). Plugging this back into the numerator, we get

\(\displaystyle x^{10}x^{-8}\), which simplifies to \(\displaystyle x^2\). Plugging this back into the original equation gives us

\(\displaystyle \frac{x^2}{x^2}\), which is simply \(\displaystyle 1\).

When like bases with exponents are multiplied, the value of the product's exponent is the sum of both original exponents as shown here:

\(\displaystyle a^{n}\cdot a^{m}=a^{n+m}\)

We can use this common rule to solve for \(\displaystyle a\) in the practice problem:

\(\displaystyle \\a^{2}\cdot a^{3}=243\\a^{2}\cdot a^{3}=a^{2+3}\\a^{5}=243\)

\(\displaystyle a=\sqrt[5]{243}\\a=3\)

The product of dividing like bases with exponents is the difference of the numerator and denominator exponents. This is a common rule when working with rational exponents:

\(\displaystyle \frac{a^{m}}{a^{n}}=a^{m-n}\)

We can use this common rule to solve for \(\displaystyle a\):

\(\displaystyle \\\frac{a^{7}}{a^{3}}=256\\\\a^{7-3}=256\\a^{4}=256\\a=\sqrt[4]{256}\\a=4\)

To solve for \(\displaystyle x\), we need all values to have like bases:

\(\displaystyle \\3^{x-2}\cdot9^{2x}=27\\3^{x-2}\cdot3^{2\cdot2x}=27\\3^{x-2}\cdot3^{4x}=3^{3}\)

Now that all values have like bases, we can solve for \(\displaystyle x\):

\(\displaystyle \\x-2+4x=3\\5x-2=3\\5x=5\\x=1\)

To solve for \(\displaystyle x\), we want all values in the equation to have like bases:

\(\displaystyle \\27^{\frac{x-3}{3}}\cdot3^{\frac{3x}{4}}=81 \\3^{3\cdot\frac{x-3}{3}}\cdot3^{\frac{3x}{4}}=3^{4} \\3^{x-3}\cdot3^{\frac{3x}{4}}=3^{4}\)

Now we can solve for \(\displaystyle x\):

\(\displaystyle \\x-3+\frac{3x}{4}=4\\\frac{7x}{4}=7\\7x=28\\x=4\)

To solve for \(\displaystyle x\), we want all the values in the equation to have like bases:

\(\displaystyle \\5^{\frac{x}{4}}\cdot5^{\frac{x}{8}}=125 \\5^{\frac{x}{4}}\cdot5^{\frac{x}{8}}=5^3\)

Now we can solve for \(\displaystyle x\):

\(\displaystyle \\\frac{x}{4}+\frac{x}{8}=3 \\8\cdot(\frac{x}{4}+\frac{x}{8})=3\cdot8 \\2x+x=24\\3x=24\\x=8\)

The original expression can be rewritten as 

\(\displaystyle (\frac{x^{\sqrt{2}}}{x^2})^2\). Whenever you can a fraction raised to a power, that power gets distributed out to the numerator and denominator. In mathematical terms, the new expression is

\(\displaystyle \frac{x^{2}}{x^4}\), which simply becomes \(\displaystyle \frac{1}{x^2}\), or \(\displaystyle x^{-2}\)