First, simplify the second expression. When there is an exponent outside the parentheses, it must be distributed to every term in the parentheses:

\(\displaystyle (2x)^3=8x^3\).

Then multiply the two terms:

\(\displaystyle 9x\cdot 8x^3=72x^4\).

In order to divide the exponents, we must have the same bases. 

Convert the base 4 to 2 by rewriting the four as two squared.

\(\displaystyle \frac{2^7}{4^{10}}=\frac{2^7}{2^{2(10)}}= \frac{2^7}{2^{20}}\)

Now that the bases are the same, subtract the power on the numerator with the denominator.

\(\displaystyle 2^{7-20}= 2^{-13} = \frac{1}{2^{13}}\)

The answer is:  \(\displaystyle \frac{1}{2^{13}}\)

When dividing exponents with the same base, we subtract the exponents while keeping the base the same.

\(\displaystyle 9^9\div9^{-9}=9^{9-(-9)}=9^{18}\)

To simplify a fraction like this, I look at the like terms separately and simplify those, putting everything together at the end.

First,

\(\displaystyle \frac{64}{80}=\frac{4}{5}\).

Then,

\(\displaystyle \frac{x^{-3}}{x^2}=\frac{1}{x^5}\).

\(\displaystyle \frac{y^{5}}{y^{-2}}=y^7\).

\(\displaystyle \frac{z^9}{z^3}=z^6\).

Put those all together to get your answer:

\(\displaystyle \frac{4y^7z^6}{5x^5}.\)

Multiply the like terms. Remember that when multiplying with exponents and bases are the same, add exponents.

Therefore,

\(\displaystyle 4(-2)=-8\)

and 

\(\displaystyle x(x^3)(x^{-2})=x^2\).

Thus, your answer is:

\(\displaystyle -8x^2.\)

When dividing variables with exponents, the exponents are subtracted

Thus

\(\displaystyle \frac{x^7y^4+x^8y^3+x^3y^7+x^3}{x^3y^3}\)

becomes

\(\displaystyle x^4y^1+x^5+y^4+y^{-3}\)

Although the bases are not the same, we know that \(\displaystyle 49=7^2\). Therefore, \(\displaystyle 49^8=(7^2)^8=7^{16}\). With the same bases, we now can subtract the exponents while keeping the base the same.

\(\displaystyle 7^{16}\div7^{-8}=7^{16-(-8)}=7^{24}\)

Notice that we can rewrite the second term in terms of the base of four.  Sixteen is equal to four squared.

\(\displaystyle 16= 4^2\)

We can replace this term with 16 in order to multiply, and then add the exponents.

\(\displaystyle 4^{30}\cdot 16^{12} =4^{30}\cdot 4^{2(12)} =4^{30}\cdot 4^{24}\)

According to the rule of exponents, whenever powers of a similar base are multiplied, the exponents can be added.

\(\displaystyle x^A \cdot x^B = x^{( A + B )}\)

The answer is:  \(\displaystyle 4^{54}\)

When dividing powers of a similar base, the exponents can be subtracted.

\(\displaystyle \frac{4^{34}}{4^{36}} = 4^{34-36}=4^{-2}\)

Simplify the negative exponent.

\(\displaystyle x^{-n} = \frac{1}{x^n}\)

\(\displaystyle 4^{-2} = \frac{1}{4^2} = \frac{1}{16}\)

The answer is:  \(\displaystyle \frac{1}{16}\)

In order to simplify this, we will need to rewrite the base 27 as a common base of three in order to be able to subtract the exponents.

\(\displaystyle 27 = 3^3\)

Rewrite the fraction.

\(\displaystyle \frac{3^{15}}{27^{18}} = \frac{3^{15}}{3^{3(18)}} = \frac{3^{15}}{3^{54}}\)

Subtract the exponents.

\(\displaystyle 3^{15-54} = 3^{-39}=\frac{1}{3^{39}}\)

The answer is:  \(\displaystyle \frac{1}{3^{39}}\)