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or, x2  3x + 1 = 0
or 3x2  9x + 3 = 0 ........(i)
Using equation (i), we get
2y^{2} – y – 3 = 0
2y^{2} – 3y + 2y – 3 = 0
y(2y – 3) + 1(2y – 3) = 0
(y + 1)(2y – 3) = 0
y = –1 or 3/2
When y = –1,
⇒ x^{2} + x = 1
x^{2} + x 1 = 0
When y = 3/2
2x^{2} – 2 = 3x
2x^{2} – 3x – 2 = 0
2x^{2} – 4x + x – 2 = 0
2x(x – 2) + 1(x – 2) = 0
(2x + 1)(x – 2) = 0
The above expression can also be written as follows:
= log_{xyz} x + log_{xyz} y + log_{xyz} z
= log_{xyz} (xyz) = 1
log_{11} 1650 = log_{11} (11 × 150) = log_{11} 11 + log_{11} 150 = 1 + log_{11} 150
Now, as 112 = 121 and as 121 < 150, we have log_{11} 150 > 2.
Thus, log_{11} 1650 > 1 + 2.
Or, log_{11} 1650 > 3
log_{13} 1950 = log_{13} (13 × 150) = log_{13} 13 + log_{13} 150 = 1 + log_{13} 150
Now, as 13^{2} = 169 and as 169 > 150, we have log_{13} 150 < 2.
Thus, log_{13} 1950 < 1 + 2 or log_{13} 1950 < 3.
⇒ log_{13} 1950 < log_{11} 1650
Hence, answer option a is correct.
If x = log24 + log35 + log46 + log57 + log68 + log79, then which of the following is true?
But log_{2}4 × log_{3}5 × ……….. × log_{7}9
= 3 × 2 = 6
∴ log24 + log35 + …………. + log79 ≥ 6 × ≥ 6^{7/6}.
If1/3 log_{3} M + 3 log_{3} N = 1 + log0.008 5, then
The given equation boils down to the following.
Hence, answer option b is correct.
If then find the value of a^{a} b^{b} c^{c}.
Let
Thus, loga = (b  c)d; logb = (c  a)d; logc = d(a  b)
Now, let a^{a}b^{b}c^{c} = f
Or, log(a^{a}b^{b}c^{c}) = logf
Or, a loga + b logb + c logc = logf
Plugging in the valued of loga, logb & logc in terms of d in the above equation,
we get
a (b  c)d + b(c  a)d + c(a  b)d = logf
or 0 = log f
or, f = 1
∴ a^{a}b^{b}c^{c} = 1
If log_{10} x  log_{10}= 2 log_{x} 10, then x is equal to
We know that log10log10 ^{x}
Therefore, the equation becomes
log10 x  log_{10} x = 2log_{x} 10
or,log_{10} x = 2log_{x} 10 … (i)
Using base change rule
Therefore, equation (i) becomes
If log (a + c), log (c  a), log (a  2b + c) are in A.P, then
log (a + c), log (c  a), log (a  2b + c) are in A.P.
log (c  a)  log (a + c) = log (a  2b + c)  log (c  a)
c^{2} + a^{2}  2ac = a^{2} + c^{2} + 2ac  2ab  2bc
4ac = 2ab + 2bc
So, a, b, c are in H. P.
Arrange the following in descending order:
A = 18^{22}, B = 22^{18}, C = 21^{17}
A = 18^{22}, B = 22^{18}, C = 21^{17}
We know that a^{b} > b^{a}, where a < b and a, b ≥ 3.
Thus, 18^{22} > 22^{18}
Moreover, 18^{22} > 22^{18} > 21^{17}
Thus, the required sequence is ABC. Hence, option (a) is correct.
The sum of the series x^{2} + + ..., x < 1, is
We know that
Replace x by x^{2}.
Find the value of x in log (x  2) + log (x  1) = log 20  1.
The given equation is equivalent to:
log (x  2)(x  1) = log 20  log 10
log (x  2)(x  1) = log 2
(x  2)(x  1)  2 = 0
x2  3x + 2  2 = 0
x(x  3) = 0
x = 0 or x = 3
x = 0 is not admissible since logarithm of a negative number (when used in the given equation) is not defined.
∴ x = 3
If log_{12 }27 = a and log9 16 = b, then find log_{8} 108.
log_{8} 108 = log_{8} (4 × 27)
log_{8} 108 = log_{8 }4 + log_{8} 27
Now,
Now
This implies:
Option (d) is the correct answer.
Thus, the expression becomes
log_{xyz} xy + log_{xyz} yz + log_{xyz} xz
= logxyz (x^{2}y^{2}z^{2}) = logxyz(xyz)^{2} = 2log_{xyz} xyz = 2
Let u = (log_{2 }x)^{2}  6 log2 x + 12, where x is a real number. Then, the equation x^{u} = 256 has
u = (log_{2} x)^{2}  6 log_{2} x + 12 & x^{u }= 256 = 2^{8} = 4^{4} = 16^{2} = 256^{1}
If x = 2 ⇒ u = (log_{2} 2)^{2}  6 log_{2} 2 + 12
u = 1  6 + 12 = 7 (Not possible)
If x = 4 ⇒ u = (log_{2} 4)^{2}  6 log_{2} 4 + 12
= (2 log_{2} 2)^{2}  12 log_{2} 2 + 12
= 4  12 + 12 = 4 (Possible)
If x = 16 ⇒ u = (log_{2} 16)2  6 log_{2} 16 + 12
u = 16  24 + 12
u = 4 (Not possible)
If x = 256 ⇒ u = (8 log_{2} 2)2  48 log_{2} 2 + 12
= 64  48 + 12
u = 28 (Not possible)
x can have exactly one solution.
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