Question

find the value of
$\frac{ {2}^{x + 3} \times {3}^{2x - y } {5}^{x + y + 3} {6}^{y + 1} }{ {6}^{x + 1} \times {10}^{y - 3} }$
is plz send me right answer only

Answer 1

$\frac{ {2}^{x + 3} \times {3}^{2x - y} \times {5}^{x + y + 3} \times {6}^{y + 1} }{ {6}^{x + 1} \times {10}^{y - 3} } \\ \\ simplify \: the \: expression\\ \\ \frac{ {2}^{x + 3} \times {3}^{2x - y} \times {5}^{x + y + 3} \times {6}^{y + 1 - (x + 1)} }{ {10}^{y - 3} } \\ \\ \frac{ {2}^{x + 3} \times {3}^{2x - y} \times {5}^{x + y + 3} \times {6}^{y - x} }{ {10}^{y - 3} }$

$factor \: the \: expression \: \\ \\ \frac{ {2}^{x + 3} \times {3}^{2x - y} \times {5}^{x + y + 3} \times {3}^{y - x} \times {2}^{y - x} }{ {10}^{y - 3} } \\ \\ calculate \: the \: product \: of \: numerator \\ \\ = > \: {2}^{x + 3 + y - x} \times {3}^{2x - y} \times {5}^{x + y + 3} \times {3}^{y - x} \\ \\ {2}^{x + 3 + y - x} \times {3}^{2x - y + y - x} \times {5}^{x + y + 3} \\ \\ {2}^{3 + y} \times {3}^{2x - x} \times {5}^{x + y + 3} \\ \\ {2}^{y + 3} \times {3}^{x} \times {5}^{x + y + 3}$

$so \: numerator \: = > \\ \\ {2}^{y + 3} \times {3}^{x} \times {5}^{x + y + 3} \\ \\ denominator \: = > \: {10}^{y - 3} \\ \\ answer \: = > \: \frac{ {2}^{y + 3} \times {3}^{x} \times {5}^{x + y + 3} }{ {10}^{y - 3} }$

Answer 2

$\sf{Given : \dfrac{(2)^x^+^3.(3)^2^x^-^y.(5)^x^+^y^+^3.(6)^y^+^1}{(6)^x^+^1.(10)^y^-^3}}$

$\bigstar \;\; \sf{We\;know\;that : \boxed{\sf{\dfrac{(a)^m}{(a)^n} = (a)^m^-^n}}}$

$\sf{\implies \dfrac{(2)^x^+^3.(3)^2^x^-^y.(5)^x^+^y^+^3.(6)^y^+^1^-^x^-^1}{(2 \times 5)^y^-^3}}$

$\sf{\implies \dfrac{(2)^x^+^3.(3)^2^x^-^y.(5)^x^+^y^+^3.(6)^y^-^x}{(2 \times 5)^y^-^3}}$

$\sf{\implies \dfrac{(2)^x^+^3.(3)^2^x^-^y.(5)^x^+^y^+^3.(2 \times 3)^y^-^x}{(2 \times 5)^y^-^3}}$

$\bigstar\;\;\sf{We\;know\;that : \boxed{\sf{(a \times b)^n = a^n\times b^n}}}$

$\sf{\implies (2 \times 3)^y^-^x = (2)^y^-^x.(3)^y^-^x}$

$\sf{\implies (2 \times 5)^y^-^3 = (2)^y^-^3.(5)^y^-^3}$

$\sf{\implies \dfrac{(2)^x^+^3.(3)^2^x^-^y.(5)^x^+^y^+^3.(2)^y^-^x.(3)^y^-^x}{(2)^y^-^3.(5)^y^-^3}}$

$\sf{\implies \dfrac{(2)^x^+^3.(2)^y^-^x.(5)^x^+^y^+^3.(3)^2^x^-^y.(3)^y^-^x}{(2)^y^-^3.(5)^y^-^3}}$

$\bigstar \;\; \sf{We\;know\;that : \boxed{\sf{(a)^m.(a)^n = (a)^m^+^n}}}$

$\sf{\implies (2)^x^+^3.(2)^y^-^x = (2)^x^+^3^+^y^-^x = (2)^y^+^3}$

$\sf{\implies (3)^2^x^-^y.(3)^y^-^x = (3)^2^x^-^y^+^y^-^x = (3)^x}$

$\sf{\implies \dfrac{(2)^y^+^3.(5)^x^+^y^+^3.(3)^x}{(2)^y^-^3.(5)^y^-^3}}$

$\sf{\implies {(2)^y^+^3^-^y^+^3.(5)^x^+^y^+^3^-^y^+^3.(3)^x}$

$\sf{\implies {(2)^6.(5)^x^+^6.(3)^x}$