Question

Simplify the following:-

$\frac{ {3}^{n} \times {9}^{n + 1} }{ {3}^{n - 1} \times {9}^{n - 1} }$

$\frac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {(25)}^{n + 1} }$

$\frac{ {5}^{n + 3} - 6 \times {5}^{n + 1} }{9 \times {5}^{x} - {2}^{2} \times {5}^{n} }$

$\frac{6 {(8)}^{n + 1} + 16 {(2)}^{3n - 2} }{10 {(2)}^{3n + 1} - 7 {(8)}^{n} }$

Please solve these questions.​

Answer 1

Answer:

4.7/5. 195. mysticd. Genius. 20.7K answers. 183.6M people helped. Hi , ( 3ⁿ × 9^n+1 )/( 3^n-1 × 9^n-1 ) = [ 3ⁿ × ( 3² )^n + 1 ]/[ ...

Step-by-step explanation:

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Answer 2

ANSWER:

$\text{1) \dfrac{3^n\times9^{n+1}}{3^{n-1}\times9^{n-1}} }\\\\:\implies\dfrac{3^n\times9^{n+1}}{3^{n-1}\times9^{n-1}}\\\\:\implies\dfrac{3^n\times3^{2(n+1)}}{3^{n-1}\times3^{2(n-1)}}\\\\:\implies\dfrac{3^n\times3^{2n+2}}{3^{n-1}\times3^{2n-2}}\\\\\text{As, a^x\times a^y=a^{x+y}\:and\:a^x\div a^y=a^{x-y} . So,}\\\\:\implies 3^{n+2n+2-(n-1)-(2n-2)}\\\\:\implies3^{n\!\!\!/\,+\,2n\!\!\!\!/\,+\,2\,-\,n\!\!\!/\,+\,1\,-\,2n\!\!\!\!/\,-\,2}\\\\:\implies 3^{1+2+2}\\\\\bf{:\implies 3^5}$

$\text{2) \dfrac{5\times25^{n+1}-25\times5^{2n}}{5\times5^{2n+3}-25^{n+1}} }\\\\:\implies\dfrac{5\times25^{n+1}-25\times5^{2n}}{5\times5^{2n+3}-25^{n+1}}\\\\:\implies\dfrac{5\times5^{2(n+1)}-5^2\times5^{2n}}{5\times5^{2n+3}-5^{2(n+1)}}\\\\:\implies\dfrac{5\times5^{2n+2}-5^2\times5^{2n}}{5\times5^{2n+3}-5^{2n+2}}\\\\\text{As, a^x\times a^y=a^{x+y} . So,}\\\\:\implies\dfrac{5^{1+2n+2}-5^{2+2n}}{5^{1+2n+3}-5^{2n+2}}\\\\:\implies\dfrac{5^{2n+3}-5^{2n+2}}{5^{2n+4}-5^{2n+2}}\\\\\text{Taking 5^{2n+2} common,}$

$:\implies\dfrac{5^{2n+2}(5-1)}{5^{2n+2}(5^2-1)}\\\\\text{ 5^{2n+2} gets cancelled. So,}\\\\:\implies\dfrac{5-1}{25-1}\\\\:\implies\dfrac{4\!\!\!/^{\:1}}{24\!\!\!\!/_{\:\:6}}\\\\\bf{:\implies\dfrac{1}{6}}$

$\text{3) \dfrac{5^{n+3}-6\times5^{n+1}}{9\times5^{n}-2^2\times5^{n}} }\\\\:\implies\dfrac{5^{n+3}-6\times5^{n+1}}{9\times5^{n}-2^2\times5^{n}}\\\\:\implies\dfrac{5^{n+3}-6\times5^{n+1}}{3^2\times5^{n}-2^2\times5^{n}}\\\\\text{Taking 5^n common.}\\\\:\implies\dfrac{5^n(5^3-6\times5)}{5^n(3^2-2^2)}\\\\\text{ 5^n gets cancelled,}\\\\:\implies\dfrac{5^3-6\times5}{3^2-2^2}\\\\:\implies\dfrac{5(25-6)}{9-4} \\\\:\implies\dfrac{5(25-6)}{5}\\\\\bf{:\implies19}$

$\text{4) \dfrac{6(8)^{n+1}+16(2)^{3n-2}}{10(2)^{3n+1}-7(8)^{n}} }\\\\:\implies\dfrac{6(8)^{n+1}+16(2)^{3n-2}}{10(2)^{3n+1}-7(8)^{n}}\\\\:\implies\dfrac{(2\times3)\times(2)^{3(n+1)}+(2^4)\times(2)^{3n-2}}{(2\times5)\times(2)^{3n+1}-(7)\times(2)^{3n}}\\\\:\implies\dfrac{2\times3\times2^{3n+3}+2^4\times2^{3n-2}}{2\times5\times2^{3n+1}-7\times2^{3n}}\\\\\text{As, a^x\times a^y=a^{x+y} . So,}\\\\:\implies\dfrac{3\times2^{1+3n+3}+2^{4+3n-2}}{5\times2^{1+3n+1}-7\times2^{3n}}$

$:\implies\dfrac{3\times2^{3n+4}+2^{3n+2}}{5\times2^{3n+2}-7\times2^{3n}}\\\\\text{Taking 2^{3n} common}\\\\:\implies \dfrac{2^{3n}(3\times2^4+2^2)}{2^{3n}(5\times2^2-7)}\\\\\text{ 2^{3n} gets cancelled,}\\\\:\implies\dfrac{3\times2^4+2^2}{5\times2^2-7}\\\\:\implies\dfrac{3\times16+4}{5\times4-7}\\\\:\implies\dfrac{48+4}{20-7}\\\\:\implies\dfrac{52\!\!\!\!/^{\:\:4}}{13\!\!\!\!/_{\:1}}\\\\\bf{:\implies 4}$

Formulae Used:

• a^x * a^y = a^{x + y}
• a^x ÷ a^y = a^{x - y}