Rewrite the second addend as follows:

$\displaystyle (9.63 \times 10^{8} ) + \left ( 8.5 \times 10^{7}\right )$

$\displaystyle = (9.63 \times 10^{8} ) + \left ( 0.85 \times 10^ {1} \times 10^{7}\right )$

$\displaystyle = (9.63 \times 10^{8} ) + \left ( 0.85 \times 10^ {1+7} \right )$

$\displaystyle = (9.63 \times 10^{8} ) + \left ( 0.85 \times 10^ {8} \right )$

$\displaystyle = (9.63 + 0.85 ) \times 10^ {8}$

$\displaystyle = 10.48 \times 10^ {8}$

This is not in scientific notation; we adjust as follows:

$\displaystyle 10.48 \times 10^ {8}$

$\displaystyle = 1.048 \times 10^{1} \times 10^ {8}$

$\displaystyle = 1.048 \times 10^ {1+ 8}$

$\displaystyle = 1.048 \times 10^ {9}$

$\displaystyle (1.5 \times 10^{-20}) \div (1.6 \times 10^{-5})$

$\displaystyle =\frac{ 1.5 \times 10^{-20} }{1.6 \times 10^{- 5}}$

$\displaystyle =\frac{ 1.5 }{1.6 } \times \frac{ 10^{-20} }{ 10^{- 5}}$

$\displaystyle =0.9375 \times 10^{-20- (-5)}$

$\displaystyle =0.9375 \times 10^{-20+5}$

$\displaystyle =0.9375 \times 10^{-15}$

Since 0.9375 does not fall in the range  $\displaystyle \left [1,10 \right )$, we adjust the answer as follows:

$\displaystyle 0.9375 \times 10^{-15}$

$\displaystyle = 9.375\times 10^{-1} \times 10^{-15}$

$\displaystyle = 9.375\times 10^{-1 + (-15)}$

$\displaystyle = 9.375\times 10^{ -16}$

Rewrite the second addend as follows:

$\displaystyle \left (7 \times 10 ^{9} \right ) - \left (7 \times 10 ^{6} \right )$

$\displaystyle = \left (7 \times 10 ^{9} \right ) - \left (0.007 \times 10 ^{9} \right )$

$\displaystyle = \left (7 - 0.007 \right ) \times 10 ^{9} \right )$

$\displaystyle =6.993 \times 10 ^{9} \right )$

$\displaystyle \left ( 8 \times 10 ^{8}\right ) ( 5 \times 10 ^{7})$

$\displaystyle = 8 \times 5 \times 10 ^{8} \times 10 ^{7}$

$\displaystyle = 40 \times 10 ^{8+7}$

$\displaystyle = 40 \times 10 ^{15 }$

Since 40 does not fall inside the rangle $\displaystyle [1,10)$, we adjust the answer as follows:

$\displaystyle 40 \times 10 ^{15 }$

$\displaystyle = 4 \times 10^{1} \times 10 ^{15 }$

$\displaystyle = 4 \times 10 ^{1+ 15 }$

$\displaystyle = 4 \times 10 ^{16 }$

We can use the fact that $\displaystyle (x^a)^b=x^{a*b}$ to see that $\displaystyle (x^2)^3=x^6.$ 

Since $\displaystyle x^2=5$, we have

 $\displaystyle (x^2)^3=(x^2)(x^2)(x^2)=(5)(5)(5)=125$.

Use the properties of exponents as follows:

$\displaystyle \left ( 2x^{2} \right )^{3} \cdot \left ( 2x^{3} \right )^{2}$

$\displaystyle = 2 ^{3} \cdot \left ( x^{2} \right )^{3} \cdot 2 ^{2} \cdot \left ( x^{3} \right )^{2}$

$\displaystyle =8 \cdot x^{2 \; \cdot \; 3} \cdot 4 \cdot x^{3 \; \cdot \; 2}$

$\displaystyle =32 \cdot x^{6} \cdot x^{6}$

$\displaystyle =32 \cdot x^{6+6}$

$\displaystyle =32x^{12}$

$\displaystyle \left (3 ^{4} \right )^{N}\cdot 3^{5} =\frac{1}{27}$

$\displaystyle 3 ^{4\cdot N} \cdot 3^{5} =3 ^{-3 }$

$\displaystyle 3 ^{4\cdot N+5} =3 ^{-3 }$

The left and right sides of the equation have the same base, so we can equate the exponents and solve:

$\displaystyle 4N + 5 = -3$

$\displaystyle 4N + 5 -5 = -3 -5$

$\displaystyle 4N = -8$

$\displaystyle 4N\div 4 = -8 \div 4$

$\displaystyle N= -2$

We'll need to remember a few logarithmic properties to answer this question:

$\displaystyle log(A\times B)=log(A)+log(B)$

$\displaystyle log(\frac{A}{B})=log(A)-log(B)$

$\displaystyle log(A^x)=x\cdot log(A)$

Now we can use these same rules to rewrite the log in question:

$\displaystyle \log \frac{100 A^{2} B}{C ^{3}}$

$\displaystyle = \log \left ( 100 A^{2} B \right )- \log \left ( C ^{3} \right )$

$\displaystyle = \log 100 + \log \left ( A^{2} \right ) + \log B - \log \left ( C ^{3} \right )$

$\displaystyle = \log 100 + 2 \log A + \log B - 3 \log C$

$\displaystyle =2 + 2 M + N - 3P$

We start by simplifying the expression on the top. Let's add the exponents inside the parentheses and then multiply by the exponent outside of the parentheses. 

Then we substract the denominator exponent form the numerator exponent:

$\displaystyle \frac{(x^{2+3})^{4}}{x^{12}}=\frac{(x^{5})^{4}}{x^{12}}=\frac{x^{5*4}}{x^{12}}=\frac{x^{20}}{x^{12}}=x^{20-12}=x^{8}$

First, we need to add the exponents of the elements with the same base that are multiplied, and subtract the exponents of same-base elements that are divided:

$\displaystyle \frac{a^7b^{11}c^{13}}{a^{13}b^{7}c^{11}}}=a^{-6}b^{4}c^{2}$

Then $\displaystyle a^{6}b^{-4}c^{-2}\cdot a^{-6}b^{4}c^{-2}=a^{0}b^{0}c^{0} or (abc)^{0}}$

Any value raised to the power of 0 equals 1 so the final result is 1.