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Given Expression
= x^{(b – c)(b + c – a)}. x^{(c – a)(c + a – b)}
. x^{(a – b)(a + b – c) }
= x^{(b – c)(b + c) – a(b – c)} . x^{(c  a)(c + a) – b(c – a)}
. x^{(a  b)(a + b)  c(a  b)}
. x^{–a(b – c) – b(c – a) – c(a – b)}
= (x^{0} × x^{0})
= (1 × 1) = 1
If m and n are whole numbers such that m^{n} = 121, the value of (m – 1)^{n + 1} is:
We know that 11^{2} = 121 = mn (given)
Hence, putting m = 11 and n = 2, in
(m – 1)^{n + 1} = (11 – 1)^{(2 + 1)} = 10^{3} = 1000
(256)^{0.16} × (256)^{0.09} = (256)^{(0.16 + 0.09)}
= (256)^{0.25}
= (256)^{(25/100)}
= (256)^{(1/4)}
= (4^{4})^{(1/4)}
= 4^{4(1/4)}
= 4^{1}
= 4
(10)1^{50} ÷ (10)^{146}
= 10^{150 – 146}
= 10^{4}
= 10000
If (25)^{7.5} × (5)^{2.5} ÷ (125)^{1.5} = 5^{x} then x = ?
Given (25)^{7.5} × (5)^{2.5} ÷ (125)^{1.5} = 5^{x}
Then, (5^{2})^{7.5} × (5)^{2.5} ÷ (5^{3})^{1.5} = 5^{x}
⇒ 5^{(2 × 7.5)} × 5^{2.5} ÷ 5^{(3 × 1.5)} = 5^{x}
⇒ 5^{15} × 5^{2.5} ÷ 5^{4.5} = 5^{x}
⇒ 5^{x} = 5^{(15 + 2.5 – 4.5)}
⇒ 5^{x} = 5^{13}
⇒ x = 13
(0.04)^{ – 1.5} = (4/100) ^{–1.5} = (1/25) ^{–3/2}
= (25)^{(3/2)}
= (5^{2})^{(3/2)}
= (5)^{2 × (3/2)}
= 5^{3}
= 125
If 3^{(x – y)} = 27 and 3^{(x + y)} = 243, then x is equal to:
3^{x – y} = 27 = 3^{3} ⇒ x – y = 3 ....(i)
3^{x + y} = 243 = 3^{5} ⇒ x + y = 5 .... (ii)
On solving (i) and (ii), we get x = 4.
If 5^{a} = 3125, then the value of 5^{(a – 3)} is:
5^{a} = 3125 ⇔ 5^{a} = 5^{5}
⇒ a = 5
∴ 5^{(a – 3)} = 5^{(5 – 3)} = 5^{2} = 25
Given that 10^{0.48 }= x, 10^{0.70} = y and x^{z} = y^{2}, then the value of z is close to:
x^{z} = y^{2}
⇒ 10^{(0.48z)} = 10^{(2 × 0.70)} = 10^{1.40}
⇒ 0.48z = 1.40
⇒ z =140/48 = 35/12 = 2.9 (approx.)
Let (17)^{3.5} × (17)^{x} = 17^{8}
Then, (17)^{3.5 + x} = 17^{8}
∴ 3.5 + x = 8
⇒ x = (8 – 3.5)
⇒ x = 4.5
= 10 × 1.732 = 17.32
(18a^{8}b^{6}) ÷ (3a^{2}b^{2}) = 18/3 × a^{8 – 2} × b^{6 – 2}
= 6a^{6}b^{4}
Replace question mark with the suitable answer:
56  45 √? = √36
56  45 √? = √36
⇒ 11  √36 =√?
⇒ 11  6 =√?
∴ ? = 25
(0.003)^{3} = 0.003 × 0.003 × 0.003
= 0.000000027
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