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lets start
\(\bullet \;\; \displaystyle \frac{2^{10}}{2^3}\)
SOLUTION: we know that \(\displaystyle \frac{a^m}{a^n} =a^{m-n} \;\;\;\) so we are going to use this here
here \(\;\;a= 2\) lets take \(\;\;m =10 ,n =3\) so by ules of exponent we have
\(\displaystyle \frac{2^{10}}{2^3}=2^{10-3}=2^7\)
\(\bullet\;\; \displaystyle \frac{5^{10} \times 2^{5}}{5^{6} \times 2^4}\)
SOLUTION: Above expression can also be written as
\( \;\; \displaystyle \frac{5^{10} \times 2^{5}}{5^{6} \times 2^4 }= \frac{5^{10}}{5^{6}} \times \frac{2^{5}}{2^4}\)
again using rules of exponents we get
\( \displaystyle \frac{5^{10}}{5^{6}} \times \frac{2^{5}}{2^4}= 5^{10-6} \times 2^{5-4}=5^{4}\times2^{1}\)
ANSWER: \(5^{4} \times 2\)
\(\bullet \;\; \displaystyle \frac{4^{5}}{4^3}\)
SOLUTION:
Above Expression can be written as
\( \displaystyle \frac{4^{5}}{4^3}= 4^{5-3}\;=\; 4^{2}\;\;\;\; \cdots\cdots 1\)
\(\text{and we know that}\;\)\(\;\; 4=2^2\) so we will use this in (1)
\(\Rightarrow \displaystyle 4^{2}\;=\; (2^2)^{2}\)
\(\Rightarrow (2^2)^{2}\;=\;2^{2 \times 2}\)
\(\Rightarrow 2^{2 \times 2} \;=\; 2^4\)
ANSWER: \(\;\;2^4 \;\;\textbf{or} \;\;16\)
\(\bullet \;\; \displaystyle \frac{2^5 \times 10^{2}}{4^2 \times 25}\)
SOLUTION:
Above Expression can be written as
\(\Rightarrow\displaystyle \frac{2^5 \times 10^{2}}{4^2 \times 25} = \frac{2^5 \times (2\times 5)^{2}}{(2^2)^2 \times 5^{2}}\)
\(\Rightarrow\displaystyle \frac{2^5 \times (2\times 5)^{2}}{(2^2)^2 \times 5^{2}} = \frac{2^5 \times 2^{2}\times 5^{2}}{2^4 \times 5^{2}}\)
\(\Rightarrow\displaystyle \frac{2^5 \times 2^{2}\times 5^{2}}{2^4 \times 5^{2}} = \frac{2^{5+2} \times 5^{2}}{2^4 \times 5^{2}}\)
\(\Rightarrow\displaystyle \frac{2^{5+2} \times 5^{2}}{2^4 \times 5^{2}} = \frac{2^{7} \times 5^{2}}{2^4 \times 5^{2}}\)
\(\Rightarrow\displaystyle \frac{2^{7} \times 5^{2}}{2^4 \times 5^{2}} = \frac{2^{7}}{2^4} \times \frac{5^{2}}{5^{2}}\)
\(\Rightarrow\displaystyle \frac{2^{7}}{2^4} \times \frac{5^{2}}{5^{2}} = 2^{7-4} \times {5^{2-2}}\)
\(\Rightarrow\displaystyle 2^{7-4} \times {5^{2-2}} = 2^{3} \times {5^{0}}\)
\(\text{we know that}\)\(\;\; a^{0} =1 \;, a \neq 0 \) using this we can write above equation as
\(\Rightarrow\displaystyle 2^{3} \times {5^{0}}=8\times 1\)
ANSWER: \(\;\;8 \)
\(\bullet \;\; \displaystyle \frac{2^{10}}{2^3}\)
SOLUTION: we know that \(\displaystyle \frac{a^m}{a^n} =a^{m-n} \;\;\;\) so we are going to use this here
here \(\;\;a= 2\) lets take \(\;\;m =10 ,n =3\) so by ules of exponent we have
\(\displaystyle \frac{2^{10}}{2^3}=2^{10-3}=2^7\)
\(\bullet\;\; \displaystyle \frac{5^{10} \times 2^{5}}{5^{6} \times 2^4}\)
SOLUTION: Above expression can also be written as
\( \;\; \displaystyle \frac{5^{10} \times 2^{5}}{5^{6} \times 2^4 }= \frac{5^{10}}{5^{6}} \times \frac{2^{5}}{2^4}\)
again using rules of exponents we get
\( \displaystyle \frac{5^{10}}{5^{6}} \times \frac{2^{5}}{2^4}= 5^{10-6} \times 2^{5-4}=5^{4}\times2^{1}\)
ANSWER: \(5^{4} \times 2\)
\(\bullet \;\; \displaystyle \frac{4^{5}}{4^3}\)
SOLUTION:
Above Expression can be written as
\( \displaystyle \frac{4^{5}}{4^3}= 4^{5-3}\;=\; 4^{2}\;\;\;\; \cdots\cdots 1\)
\(\text{and we know that}\;\)\(\;\; 4=2^2\) so we will use this in (1)
\(\Rightarrow \displaystyle 4^{2}\;=\; (2^2)^{2}\)
\(\Rightarrow (2^2)^{2}\;=\;2^{2 \times 2}\)
\(\Rightarrow 2^{2 \times 2} \;=\; 2^4\)
ANSWER: \(\;\;2^4 \;\;\textbf{or} \;\;16\)
\(\bullet \;\; \displaystyle \frac{2^5 \times 10^{2}}{4^2 \times 25}\)
SOLUTION:
Above Expression can be written as
\(\Rightarrow\displaystyle \frac{2^5 \times 10^{2}}{4^2 \times 25} = \frac{2^5 \times (2\times 5)^{2}}{(2^2)^2 \times 5^{2}}\)
\(\Rightarrow\displaystyle \frac{2^5 \times (2\times 5)^{2}}{(2^2)^2 \times 5^{2}} = \frac{2^5 \times 2^{2}\times 5^{2}}{2^4 \times 5^{2}}\)
\(\Rightarrow\displaystyle \frac{2^5 \times 2^{2}\times 5^{2}}{2^4 \times 5^{2}} = \frac{2^{5+2} \times 5^{2}}{2^4 \times 5^{2}}\)
\(\Rightarrow\displaystyle \frac{2^{5+2} \times 5^{2}}{2^4 \times 5^{2}} = \frac{2^{7} \times 5^{2}}{2^4 \times 5^{2}}\)
\(\Rightarrow\displaystyle \frac{2^{7} \times 5^{2}}{2^4 \times 5^{2}} = \frac{2^{7}}{2^4} \times \frac{5^{2}}{5^{2}}\)
\(\Rightarrow\displaystyle \frac{2^{7}}{2^4} \times \frac{5^{2}}{5^{2}} = 2^{7-4} \times {5^{2-2}}\)
\(\Rightarrow\displaystyle 2^{7-4} \times {5^{2-2}} = 2^{3} \times {5^{0}}\)
\(\text{we know that}\)\(\;\; a^{0} =1 \;, a \neq 0 \) using this we can write above equation as
\(\Rightarrow\displaystyle 2^{3} \times {5^{0}}=8\times 1\)
ANSWER: \(\;\;8 \)
\(\bullet \;\; \displaystyle \frac{2^2 \times 3^{2}}{18^2 \times 3^5\times 5^2}\)
SOLUTION:
Above Expression can be written as
\( \displaystyle \frac{2^2 \times 3^{2}}{18^2 \times 3^5\times 5^2}=\frac{2^2 \times 3^{2}}{(2 \times 9)^2 \times 3^5\times 5^2}\)
\(\Rightarrow\displaystyle\frac{2^2 \times 3^{2}}{(2 \times 3^2)^2 \times 3^5\times 5^2}\)
\(\Rightarrow\displaystyle \frac{2^2 \times 3^{2}}{2^2 \times (3^2)^2 \times 3^5\times 5^2}\)
\(\Rightarrow\displaystyle \frac{2^2 \times 3^{2}}{2^2 \times 3^4 \times 3^5\times 5^2}\)
\(\Rightarrow\displaystyle \frac{2^2 \times 3^{2}}{2^2 \times 3^{4+5} \times 5^2}\)
\(\Rightarrow\displaystyle \frac{2^2 \times 3^{2}}{2^2 \times 3^{9} \times 5^2}\)
now above expression can be written as
\(\Rightarrow\displaystyle \frac{2^2}{2^2} \times \frac{3^2}{3^9}\times \frac{1}{5^2}\)
\(\Rightarrow\displaystyle \frac{2^{2-2} \times 3^{2-9} }{5^2}\)
\(\Rightarrow\displaystyle \frac{2^{0}\times 3^{-7}}{5^2}\)
\(\Rightarrow\displaystyle \frac{2^{0}\times 3^{-7}}{5^2}\)
We know form exponents formulas \(\;\; \displaystyle a^{-m}=\frac{1}{a^m} \) or \(\displaystyle\;\;a^m = \frac{1}{a^{-m}}\)
We will use above,So we will get
\(\Rightarrow \displaystyle \frac{1}{3^7 \times 5^2}\)
ANSWER: \(\;\; \displaystyle \frac{1}{ 3^{7}\times {5^2}}\)
\(\bullet \;\; \displaystyle (-2)^5 \times (-3)^{10}\)
SOLUTION: we can write given expression as
\((-2)^5 \times (-3)^{10}=(-1 \times 2)^5 \times (-1 \times 3)^{10}\)
\(\Rightarrow \displaystyle (-1)^5 \times (2)^5 \times (-1)^{10} \times (3)^{10} \;\;\;\cdots \cdots (1) \)
We have rule that
\(\;\;(-1)^n = \begin{cases} -1 & \text{n is odd} \\ 1 & \text{n is even} \end{cases}\)
So form above rule (1) will become
\(\Rightarrow \displaystyle (-1) \times 2^5 \times 1 \times 3^{10} \)
\(\Rightarrow \displaystyle -2^5 \times 3^{10} \)
ANSWER: \(\;\; -2^5 \times 3^{10}\)
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