Test Level 2: Exponents and Logarithm - 2 - CAT MCQ

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

Using equation (i), we get

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

The above expression can also be written as follows:


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

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.

But log₂4 × log₃5 × ……….. × log₇9

= 3 × 2 = 6
∴ log24 + log35 + …………. + log79 ≥ 6 ×  ≥ 6^7/6.

The given equation boils down to the following.

Hence, answer option b is correct.

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

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

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.

A = 18²², B = 22¹⁸, C = 21¹⁷
We know that a^b > b^a, 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.

We know that

Replace x by -x².

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

 log₈ 108 = log₈ (4 × 27)
log₈ 108 = log₈ 4 + log₈ 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 (x²y²z²) = logxyz(xyz)² = 2log_xyz xyz = 2

u = (log₂ x)² - 6 log₂ x + 12 & x^u = 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.