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maths SM ch1 pg1.35 Q3,4,5,6

Mark BRAINIEST only to ans. giving explaining correctly DETAILED ANS. of ALL Q.s
​

Answer 1

$\mathsf{3.\;\;Given : \dfrac{(2)^{x + 3} \times (3)^{2x - y} \times (5)^{x + y + 3} \times (6)^{y + 1}}{(6)^{x + 1} \times (10)^{y + 3} \times (15)^x}}$

★  6 can be written as : (2 × 3)

★  10 can be written as : (5 × 2)

★  15 can be written as : (5 × 3)

$\mathsf{\implies \dfrac{(2)^{x + 3} \times (3)^{2x - y} \times (5)^{x + y + 3} \times \big(2 \times3\big)^{y + 1}}{\big(2 \times 3\big)^{x + 1} \times \big(5 \times 2\big)^{y + 3} \times \big(5 \times 3\big)^x}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{\big(ab\big)^n = (a)^n(b)^n}}}$

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

$\mathsf{\implies \dfrac{\big[(2)^{x + 3} \times (2)^{y + 1}\big] \big[(3)^{2x - y} \times (3)^{y + 1}\big] \big[(5)^{x + y + 3}\big]}{\big[(2)^{x + 1} \times (2)^{y + 3}\big] \big[(3)^{x + 1} \times (3)^x\big] \big[(5)^{y + 3} \times (5)^x\big]}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{(a)^m \times(a)^n = (a)^{m + n}}}}$

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

$\mathsf{\implies \dfrac{(2)^{(x + y + 4)} \times (3)^{(2x + 1)}}{(2)^{(x + y + 4)} \times (3)^{(2x + 1)}}}$

$\mathsf{\implies 1}$

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$\mathsf{4.\;\;Given :\;\dfrac{(9)^y.(3)^2.\big(3^{-y}\big)^{-1} - (27)^y}{(3)^{3x}.(2)^{3}} = \dfrac{1}{27}}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{\big[(a)^m\big]^n = (a)^{mn}}}}$

$\mathsf{\implies \dfrac{(9)^y.(3)^2.(3)^y - (27)^y}{(3^3)^{x}.(8)} = \dfrac{1}{27}}}$

$\mathsf{\implies \dfrac{(9)^y.(3)^y(3)^2 - (27)^{y}}{(27)^{x}.(8)} = \dfrac{1}{27}}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{(a)^n(b)^n = \big(ab\big)^n}}}$

$\mathsf{\implies \dfrac{(9 \times 3)^y.(3)^2 - (27)^{y}}{(27)^{x}.(8)} = \dfrac{1}{27}}}$

$\mathsf{\implies \dfrac{(27)^y.(3)^2 - (27)^{y}}{(27)^{x}.(8)} = \dfrac{1}{27}}}$

$\mathsf{\implies \dfrac{(27)^y\big[(3)^2 - 1\big]}{(27)^{x}.(8)} = \dfrac{1}{27}}}$

$\mathsf{\implies \dfrac{(27)^y\big[9 - 1\big]}{(27)^{x}.(8)} = \dfrac{1}{27}}}$

$\mathsf{\implies \dfrac{(27)^y(8)}{(27)^{x}.(8)} = \dfrac{1}{27}}}$

$\mathsf{\implies \dfrac{(27)^y}{(27)^{x}} = \dfrac{1}{27}}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{\dfrac{a^m}{a^n} = a^{m - n}}}}$

$\mathsf{\implies (27)^{y - x} = 27^{-1}}$

★  When Bases are same - Exponents should be equal

$\mathsf{\implies y - x = -1}$

$\mathsf{\implies x - y = 1}$

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$\mathsf{5.\;\;Given : \;\bigg[(x)^{\dfrac{1}{a - b}}\bigg]^{\dfrac{1}{a - c}} \times \bigg[(x)^{\dfrac{1}{b - c}}\bigg]^{\dfrac{1}{b - a}} \times \bigg[(x)^{\dfrac{1}{c - a}}\bigg]^{\dfrac{1}{c - b}}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{\big[(a)^m\big]^n = (a)^{mn}}}}$

$\mathsf{\implies (x)^{\bigg[\dfrac{1}{(a - b)(a - c)}\bigg]} * (x)^{\bigg[\dfrac{1}{(b - c)(b - a)}\bigg]} * (x)^{\bigg[\dfrac{1}{(c - a)(c - b)}\bigg]}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{(a)^m \times(a)^n = (a)^{m + n}}}}$

$\mathsf{\implies (x)^{\bigg[\dfrac{1}{(a - b)(a - c)} + \dfrac{1}{(b - c)(b - a)} + \dfrac{1}{(c - a)(c - b)}\bigg]}}}$

$\mathsf{\implies (x)^{\bigg[\dfrac{1}{(a - b)(a - c)} - \dfrac{1}{(b - c)(a - b)} + \dfrac{1}{(c - a)(c - b)}\bigg]}}}$

$\mathsf{\implies (x)^{\bigg[\dfrac{1}{a - b}\bigg(\dfrac{1}{a - c} - \dfrac{1}{b - c}\bigg) + \dfrac{1}{(c - a)(c - b)}\bigg]}}}$

$\mathsf{\implies (x)^{\bigg[\dfrac{1}{a - b}\bigg(\dfrac{(b - c) - (a - c)}{(a - c)(b - c)}\bigg) + \dfrac{1}{(c - a)(c - b)}\bigg]}}}$

$\mathsf{\implies (x)^{\bigg[\dfrac{1}{a - b}\bigg(\dfrac{(b - c - a + c)}{(a - c)(b - c)}\bigg) + \dfrac{1}{(c - a)(c - b)}\bigg]}}}$

$\mathsf{\implies (x)^{\bigg[\dfrac{1}{a - b}\bigg(\dfrac{(b - a)}{(a - c)(b - c)}\bigg) + \dfrac{1}{(c - a)(c - b)}\bigg]}}}$

$\mathsf{\implies (x)^{\bigg[\dfrac{1}{a - b}\bigg(\dfrac{-(a - b)}{(a - c)(b - c)}\bigg) + \dfrac{1}{(c - a)(c - b)}\bigg]}}}$

$\mathsf{\implies (x)^{\bigg[\dfrac{-1}{(a - c)(b - c)} + \dfrac{1}{(c - a)(c - b)}\bigg]}}}$

$\mathsf{\implies (x)^{\bigg[\dfrac{1}{(a - c)(c - b)} + \dfrac{1}{(c - a)(c - b)}\bigg]}}}$

$\mathsf{\implies (x)^{\bigg[\dfrac{-1}{(c - a)(c - b)} + \dfrac{1}{(c - a)(c - b)}\bigg]}}}$

$\mathsf{\implies (x)^{0}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{a^0 = 1}}}$

$\mathsf{\implies (x)^{0} = 1}$

Answer 2

${\orange{\huge{\mathfrak{answer}}}}$

★ 6 can be written as : (2 × 3)

★ 10 can be written as : (5 × 2)

★ 15 can be written as : (5 × 3)

$\mathsf{\implies \dfrac{(2)^{x + 3} \times (3)^{2x - y} \times (5)^{x + y + 3} \times \big(2 \times3\big)^{y + 1}}{\big(2 \times 3\big)^{x + 1} \times \big(5 \times 2\big)^{y + 3} \times \big(5 \times 3\big)^x}}$⟹

(2×3)

x+1

×(5×2)

y+3

×(5×3)

x

(2)

x+3

×(3)

2x−y

×(5)

x+y+3

×(2×3)

y+1

$\mathsf{\implies \dfrac{(2)^{x + 3} \times (3)^{2x - y} \times (5)^{x + y + 3} \times (2)^{y + 1} \times(3)^{y + 1}}{(2)^{x + 1} \times (3)^{x + 1} \times (5)^{y + 3} \times (2)^{y + 3} \times (5)^x \times (3)^x}}$⟹

(2)

x+1

×(3)

x+1

×(5)

y+3

×(2)

y+3

×(5)

x

×(3)

x

(2)

x+3

×(3)

2x−y

×(5)

x+y+3

×(2)

y+1

×(3)

y+1

$\mathsf{\implies \dfrac{\big[(2)^{x + 3} \times (2)^{y + 1}\big] \big[(3)^{2x - y} \times (3)^{y + 1}\big] \big[(5)^{x + y + 3}\big]}{\big[(2)^{x + 1} \times (2)^{y + 3}\big] \big[(3)^{x + 1} \times (3)^x\big] \big[(5)^{y + 3} \times (5)^x\big]}}$⟹

[(2)

x+1

×(2)

y+3

][(3)

x+1

×(3)

x

][(5)

y+3

×(5)

x

]

[(2)

x+3

×(2)

y+1

][(3)

2x−y

×(3)

y+1

][(5)

x+y+3

]

$\mathsf{\implies \dfrac{(2)^{(x + 3 + y + 1)} \times (3)^{(2x - y + y + 1)} \times (5)^{(x + y + 3)}}{(2)^{(x + 1 + y + 3)} \times (3)^{(x + 1 + x)} \times (5)^{(x + y + 3)}}}$⟹

(2)

(x+1+y+3)

×(3)

(x+1+x)

×(5)

(x+y+3)

(2)

(x+3+y+1)

×(3)

(2x−y+y+1)

×(5)

(x+y+3)

$\mathsf{\implies \dfrac{(2)^{(x + y + 4)} \times (3)^{(2x + 1)}}{(2)^{(x + y + 4)} \times (3)^{(2x + 1)}}}$⟹

(2)

(x+y+4)

×(3)

(2x+1)

(2)

(x+y+4)

×(3)

(2x+1)

$\mathsf{\implies 1}$⟹1

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$\mathsf{\implies (27)^{y - x} = 27^{-1}}$⟹(27)

y−x

=27

−1

★ When Bases are same - Exponents should be equal

$\mathsf{\implies y - x = -1}$⟹y−x=−1

$\mathsf{\implies x - y = 1}$⟹x−y=1

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$\mathsf{5.\;\;Given : \;\bigg[(x)^{\dfrac{1}{a - b}}\bigg]^{\dfrac{1}{a - c}} \times \bigg[(x)^{\dfrac{1}{b - c}}\bigg]^{\dfrac{1}{b - a}} \times \bigg[(x)^{\dfrac{1}{c - a}}\bigg]^{\dfrac{1}{c - b}}}$5.Given:[(x)

a−b

1

]

a−c

1

×[(x)

b−c

1

]

b−a

1

×[(x)

c−a

1

]

c−b

1

$\mathsf{\implies (x)^{\bigg[\dfrac{1}{(a - b)(a - c)}\bigg]} * (x)^{\bigg[\dfrac{1}{(b - c)(b - a)}\bigg]} * (x)^{\bigg[\dfrac{1}{(c - a)(c - b)}\bigg]}}$⟹(x)

[

(a−b)(a−c)

1

]

∗(x)

[

(b−c)(b−a)

1

]

∗(x)

[

(c−a)(c−b)

1

]

$\mathsf{\implies (x)^{0}}[/te ]⟹(x) </p><p>0</p><p> </p><p></p><p></p><p></p><p>[tex]\mathsf{\implies (x)^{0} = 1}$⟹(x)

0

=1

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