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Test Level 2: Exponents and Logarithm - 2


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Test Level 2: Exponents and Logarithm - 2 - Question 1

If x + 1/x = 3, then the value of is

Detailed Solution for Test Level 2: Exponents and Logarithm - 2 - Question 1


or, x2 - 3x + 1 = 0
or 3x2 - 9x + 3 = 0 ........(i)

Using equation (i), we get

Test Level 2: Exponents and Logarithm - 2 - Question 2

If – 7 = 0, then two values of x are

Detailed Solution for Test Level 2: Exponents and Logarithm - 2 - Question 2




2y2 – y – 3 = 0
2y2 – 3y + 2y – 3 = 0
y(2y – 3) + 1(2y – 3) = 0
(y + 1)(2y – 3) = 0
y = –1 or 3/2
When y = –1,
⇒ x2 + x = 1
x2 + x -1 = 0

When y = 3/2

2x2 – 2 = 3x
2x2 – 3x – 2 = 0
2x2 – 4x + x – 2 = 0
2x(x – 2) + 1(x – 2) = 0
(2x + 1)(x – 2) = 0

Test Level 2: Exponents and Logarithm - 2 - Question 3

Find the value of.

Detailed Solution for Test Level 2: Exponents and Logarithm - 2 - Question 3


The above expression can also be written as follows:


= logxyz x + logxyz y + logxyz z
= logxyz (xyz) = 1

Test Level 2: Exponents and Logarithm - 2 - Question 4

Which of the following is true?

Detailed Solution for Test Level 2: Exponents and Logarithm - 2 - Question 4

log11 1650 = log11 (11 × 150) = log11 11 + log11 150 = 1 + log11 150
Now, as 112 = 121 and as 121 < 150, we have log11 150 > 2.
Thus, log11 1650 > 1 + 2.
Or, log11 1650 > 3
log13 1950 = log13 (13 × 150) = log13 13 + log13 150 = 1 + log13 150

Now, as 132 = 169 and as 169 > 150, we have log13 150 < 2.
Thus, log13 1950 < 1 + 2 or log13 1950 < 3.
⇒ log13 1950 < log11 1650
Hence, answer option a is correct.

Test Level 2: Exponents and Logarithm - 2 - Question 5

If x = log24 + log35 + log46 + log57 + log68 + log79, then which of the following is true?

Detailed Solution for Test Level 2: Exponents and Logarithm - 2 - Question 5


But log24 × log35 × ……….. × log79

= 3 × 2 = 6
∴ log24 + log35 + …………. + log79 ≥ 6 ×  ≥ 67/6.

Test Level 2: Exponents and Logarithm - 2 - Question 6

If1/3 log3 M + 3 log3 N = 1 + log0.008 5, then

Detailed Solution for Test Level 2: Exponents and Logarithm - 2 - Question 6


The given equation boils down to the following.

Hence, answer option b is correct.

Test Level 2: Exponents and Logarithm - 2 - Question 7

If then find the value of aa bb cc.

Detailed Solution for Test Level 2: Exponents and Logarithm - 2 - Question 7

Let 
Thus, loga = (b - c)d; logb = (c - a)d; logc = d(a - b)
Now, let aabbcc = f
Or, log(aabbcc) = 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
∴ aabbcc = 1

Test Level 2: Exponents and Logarithm - 2 - Question 8

If log10 x - log10= 2 logx 10, then x is equal to

Detailed Solution for Test Level 2: Exponents and Logarithm - 2 - Question 8

We know that log10log10 x
Therefore, the equation becomes
log10 x - log10 x = 2logx 10
or,log10 x = 2logx 10 … (i)
Using base change rule 
Therefore, equation (i) becomes

Test Level 2: Exponents and Logarithm - 2 - Question 9

If log (a + c), log (c - a), log (a - 2b + c) are in A.P, then

Detailed Solution for Test Level 2: Exponents and Logarithm - 2 - Question 9

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)

c2 + a2 - 2ac = a2 + c2 + 2ac - 2ab - 2bc
4ac = 2ab + 2bc

So, a, b, c are in H. P.

Test Level 2: Exponents and Logarithm - 2 - Question 10

Arrange the following in descending order:

A = 1822, B = 2218, C = 2117

Detailed Solution for Test Level 2: Exponents and Logarithm - 2 - Question 10

A = 1822, B = 2218, C = 2117
We know that ab > ba, where a < b and a, b ≥ 3.
Thus, 1822 > 2218
Moreover, 1822 > 2218 > 2117
Thus, the required sequence is ABC. Hence, option (a) is correct.

Test Level 2: Exponents and Logarithm - 2 - Question 11

The sum of the series x2+ ..., |x| < 1, is

Detailed Solution for Test Level 2: Exponents and Logarithm - 2 - Question 11


We know that

Replace x by -x2.

Test Level 2: Exponents and Logarithm - 2 - Question 12

Find the value of x in log (x - 2) + log (x - 1) = log 20 - 1.

Detailed Solution for Test Level 2: Exponents and Logarithm - 2 - Question 12

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

Test Level 2: Exponents and Logarithm - 2 - Question 13

If log12 27 = a and log9 16 = b, then find log8 108.

Detailed Solution for Test Level 2: Exponents and Logarithm - 2 - Question 13

 log8 108 = log8 (4 × 27)
log8 108 = log8 4 + log8 27
Now, 
Now

This implies:

Option (d) is the correct answer.

Test Level 2: Exponents and Logarithm - 2 - Question 14

equals

Detailed Solution for Test Level 2: Exponents and Logarithm - 2 - Question 14



Thus, the expression becomes
logxyz xy + logxyz yz + logxyz xz
= logxyz (x2y2z2) = logxyz(xyz)2 = 2logxyz xyz = 2

Test Level 2: Exponents and Logarithm - 2 - Question 15

Let u = (log2 x)2 - 6 log2 x + 12, where x is a real number. Then, the equation xu = 256 has

Detailed Solution for Test Level 2: Exponents and Logarithm - 2 - Question 15

u = (log2 x)2 - 6 log2 x + 12 & xu = 256 = 28 = 44 = 162 = 2561
If x = 2 ⇒ u = (log2 2)2 - 6 log2 2 + 12
u = 1 - 6 + 12 = 7 (Not possible)
If x = 4 ⇒ u = (log2 4)2 - 6 log2 4 + 12
= (2 log2 2)2 - 12 log2 2 + 12
= 4 - 12 + 12 = 4 (Possible)
If x = 16 ⇒ u = (log2 16)2 - 6 log2 16 + 12
u = 16 - 24 + 12
u = 4 (Not possible)
If x = 256 ⇒ u = (8 log2 2)2 - 48 log2 2 + 12
= 64 - 48 + 12
u = 28 (Not possible)
x can have exactly one solution.

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