When you multiply exponents, you add the common bases:

y⁴ x⁷ + y⁵x⁶ + y¹⁸x⁴ + y³x²⁶

By using exponent rules, we can simplify Quantity B.  

^{$\displaystyle \dpi{100} \small \dpi{100} \small 2^{4}\times (8^{n+1})^{2}\div 4^{n}$}

$\displaystyle \dpi{100} \small \dpi{100} \small 2^{4}\times (8^{2n+2})\div 4^{n}$

  $\displaystyle \dpi{100} \small \dpi{100} \small 2^{4}\times 2^{3(2n+2)}\div 4^{n}$

$\displaystyle \dpi{100} \small \dpi{100} \small 2^{4}\times 2^{6n+6}\div 4^{n}$ 

$\displaystyle \dpi{100} \small \dpi{100} \small 2^{6n+10}\div 4^{n}$

$\displaystyle \dpi{100} \small \dpi{100} \small 2^{6n+10}\div 2^{2n}$

$\displaystyle \dpi{100} \small 2^{4n+10}$

Also, we can simplify Quantity A. 

^{$\displaystyle \dpi{100} \small 16^{n+2}$}

$\displaystyle \dpi{100} \small =2^{4(n+2)}$

$\displaystyle \dpi{100} \small =2^{4n+8}$

Since n is positive, $\displaystyle \dpi{100} \small 4n+10>4n+8$

Rewrite the term on the left as a product. Remember that negative exponents shift their position in a fraction (denominator to numerator).

$\displaystyle 3^{y-1}*3^2=27^y3^y$

The term on the right can be rewritten, as 27 is equal to 3 to the third power.

$\displaystyle 3^{y-1}*3^2=(3^3)^y*3^y$

Exponent rules dictate that multiplying terms allows us to add their exponents, while one term raised to another allows us to multiply exponents.

$\displaystyle 3^{(y-1)+2}=(3)^{3y}*3^y$

$\displaystyle 3^{y+1}=3^{3y+y}=3^{4y}$

We now know that the exponents must be equal, and can solve for $\displaystyle y$.

$\displaystyle y+1=4y$

$\displaystyle 1=3y$

$\displaystyle \frac{1}{3}=y$

Since the base is 5 for each term, we can say 2 + n =12.  Solve the equation for n by subtracting 2 from both sides to get n = 10.

Start by simplifying each individual term between the plus signs. We can add the exponents in $\displaystyle x^3x^6$ and $\displaystyle x^3x^2x^2x^2$ so each of those terms becomes $\displaystyle x^9$. Then multiply the exponents in $\displaystyle (x^3)^3$ so that term also becomes $\displaystyle x^9$. Thus, we have simplified the expression to $\displaystyle x^9 + x^9 + x^9$ which is $\displaystyle 3x^{9}$.

First, simplify $\displaystyle x^{3}x^{5}$ by adding the exponents to get $\displaystyle x^{8}$.

Then simplify $\displaystyle (y^{2})^{2}$ by multiplying the exponents to get $\displaystyle y^{4}$.

This gives us $\displaystyle x^{8} - x^{^{2}} + y^{4}$. We cannot simplify any further.

To attempt this problem, note that $\displaystyle 3^2=(3)^2=9^1=9$.

Now note that when multiplying numbers, if the base is the same, we may add the exponents:

$\displaystyle 3^4 \cdot 3^8 = 3^{4+8}=3^{12}$

This can in turn be written in terms of nine as follows (recall above)

$\displaystyle 3^{12}=9^{\frac{12}{2}}=9^6$

$\displaystyle 9^6=9^x$

$\displaystyle x=6$

When dealing with exponenents, when multiplying two like bases together, add their exponents:

$\displaystyle 3^2\cdot3^5=3^7$

However, when an exponent appears outside of a parenthesis, or if the entire number itself is being raised by a power, multiply:

$\displaystyle (3^7)^3=3^{7\cdot 3}=3^{21}$

$\displaystyle 3^{21}=3^{x}$

$\displaystyle x=21$