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Simplify and express each of the following in exponential form:
$(i)$. $\frac{2^3\times3^4\times4}{3\times32}$
$(ii)$. $[(5^2)^{3\ }\times5^4]\div5^7$
$(iii)$. $(25^4\div5^3)$
$(iv)$. $\frac{3\times7^2\times11^8}{21\times11^3}$
$(v)$. $\frac{3^7}{3^4\times3^3}$
$(vi)$. $2^0+3^0+4^0$
$(viii)$. $2^0\times3^0\times4^0$
$(viii)$. $(3^0+2^0)\times5^0$
$(ix)$. $\frac{2^8\times a^5}{4^3\times a3}$
$(x)$. $(\frac{a^5}{a^3})\times a^8$
$(xi)$. $\frac{4^5\times a^8b^3}{4^5\times a^5b^2}$
$(xii)$. $(2^3\times2)^2$
Given:
$(i)$. $\frac{2^3\times3^4\times4}{3\times32}$
$(ii)$. $[(5^2)^{3\ }\times5^4]\div5^7$
$(iii)$. $(25^4\div5^3)$
$(iv)$. $\frac{3\times7^2\times11^8}{21\times11^3}$
$(v)$. $\frac{3^7}{3^4\times3^3}$
$(vi)$. $2^0+3^0+4^0$
$(vii)$. $2^0\times3^0\times4^0$
$(viii)$. $(3^0+2^0)\times5^0$
$(ix)$. $\frac{2^8\times a^5}{4^3\times a^3}$
$(x)$. $(\frac{a^5}{a^3})\times a^8$
$(xi)$. $\frac{4^5\times a^8b^3}{4^5\times a^5b^2}$
$(xii)$. $(2^3\times2)^2$
To do: To simplify and express each of the above in exponential form.
Solution:
$(i)$. $\frac{2^3\times3^4\times4}{3\times32}$
$=\frac{2^3\times3^4\times2^2}{3\times2^5}$
$=\frac{2^{3+2}\times3^4}{3\times2^5}$ $[\because\ a^m\times a^n=a^{m+n}]$
$=\frac{2^5\times3^4}{3\times2^5}$
$=2^{5-5}\times 3^{4-1}$
$=2^0\times3^3$ $[a^m\div a^n=a^{m-n}]$
$=1\times3^3$ $[\because a^0=1]$
$=3^3$
$(ii)$. $[(5^2)^{3\ }\times5^4]\div5^7$
$=(5^6\times5^4)\div5^7$ $[(a^m)^n=a^{m\times n}]$
$=(5^{6+4})\div5^7$ $[\because\ a^m\times a^n=a^{m+n}]$
$=5^{10}\div5^7$
$=5^{10-7}$ $[a^m\div a^n=a^{m-n}]$
$=5^3$
$(iii)$. $(25^4\div5^3)$
$=(5^2)^4\div5^3$
$=5^8\div5^3$ $[(a^m)^n=a^{m\times n}]$
$=5^{8-3}$ $[a^m\div a^n=a^{m-n}]$
$=5^5$
$(iv)$. $\frac{3\times7^2\times11^8}{21\times11^3}$
$=\frac{3\times7^2\times11^8}{3\times7\times11^3}$
$=3^{1-1}\times7^{2-1}\times11^{8-3}$ $[a^m\div a^n=a^{m-n}]$
$=3^0\times7^1\times11^5$
$=7\times11^5$
$(v)$. $\frac{3^7}{3^4\times3^3}$
$=\frac{3^7}{3^{4+3}}$ $[\because\ a^m\times a^n=a^{m+n}]$
$=\frac{3^7}{3^7}$
$=3^{7-7}$ $[a^m\div a^n=a^{m-n}]$
$=3^0$
$=1$ $[\because a^0=1]$
$(vi)$. $2^0+3^0+4^0$
$=1+1+1$ $[\because a^0=1]$
$=3$
$(vii)$. $2^0\times3^0\times4^0$
$=1\times1\times1$ $[\because a^0=1]$
$=1$
$(viii)$. $(3^0+2^0)\times5^0$
$(1+1)\times1$ $[a^0=1]$
$=2\times1$
$=2$
$(ix)$. $\frac{2^8\times a^5}{4^3\times a^3}$
$\frac{2^8\times a^5}{(2^2)^3\times a^3}$
$=\frac{2^8\times a^5}{2^6\times a^3}$
$=2^{8-2}\times a^{5-3}$ $[a^m\div a^n=a^{m-n}]$
$=2^2\times a^2$
$=(2a)^2$
$(x)$. $(\frac{a^5}{a^3})\times a^8$
$=(a^{5-3})\times a^8$ $[a^m\div a^n=a^{m-n}]$
$=(a^2)\times a^8$
$=a^{2+8}$ $[\because\ a^m\times a^n=a^{m+n}]$
$=a^{10}$
$(xi)$. $\frac{4^5\times a^8b^3}{4^5\times a^5b^2}$
$=4^{5-5}\times a^{8-5}b^{3-2}$ $[a^m\div a^n=a^{m-n}]$
$=4^0\times a^3\times b^{1}$
$=1\times a^3\times b$ $[a^0=1]$
$=a^3b$
$(xii)$. $(2^3\times2)^2$
$=(2^{3+1})^2$ $[\because\ a^m\times a^n=a^{m+n}]$
$=(2^4)^2$
$=2^{4\times 2}$ $[(a^m)^n=a^{m\times n}]$
$=2^8$
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