# Aptitude Logarithm Test Paper 2

6) Add the equation x = 1 + loga bc, y = 1 + logb ca and z = 1 + logc ab. Find the value of .

1. xyz = xy + yz + zx
2. X2yz = xyz + yx + 1
3. (xz + 1)2 = xy - yz - zx
4. (xz + y) = xy + yz + zx

Explanation:

x =1 + loga bc

We can write it as x = loga a + loga bc => loga abc

Similarly, y = logb b + logb ca => logb abc

And, z = logc c + logc ab => logc abc
Changing the values of x, y, z to a common base m by using logb a = , we get
X = Y = Z = Now, = [logm a + logm b + logm c]

Now, Right hand side equation, [logm abc/ logm abc] = 1
Now, = 1

Or, = 1 => xyz =xy + yz + zx.

7) Find the characteristics of the logarithms of i) 5631, ii) 5.678, iii) 56.23.

1. 3, 1, 5
2. 3, 0, 1
3. 6, 5, 9
4. 8, 7, 6

Explanation:

i) The number 5631> 1 and the number of digits in the integral part are 4.
Therefore, the characteristics of the logarithm are 4-1 = 3,

ii) In the number 5.678, the number of digits in the integral part = 1.
Therefore, characteristics of the logarithm are 1-1 =0

iii) In this case, the number of digits in an integral part of the number 56.23 are 2.
Therefore, the characteristics of the logarithm = 2-1 = 1

So, the option b is correct.

8) If log 2 = 0.30103, log 3 = 0.47712, the number of digits in 620 is

1. 8
2. 12
3. 16
4. 20

Explanation:

We have to take 620 with log
We know that log mn = n log m
Now, log 620 = 20 log 6
Or, 20 log (2*3)
We know that log (m*n) = log m + log n
Now, 20 [log 2 + log 3] = 20 [0.30103 + 0.47712] = 20 [0.77815]
Since, the number of digits in 620 = 15.563, round up value = 16
Therefore the number of digits = 16

9) The logarithm of 0.0001 to the base 0.001 is equal to

1. 4/3
2. 5/3
3. 7/3
4. 2/3

Explanation:

Let Log0.001 (0.0001) = x

We know that logx y = a equals to xa = y.

So, (0.001)x= 0.0001
Or, = [1/10000]

Or, = Therefore, compare both side, we get 3x = 4.

Hence, x = 4/3

10) If logb x = ͞5.1342618, then the value of log10(x(1/4)) will be

1. ͞1.2835655
2. ͞2.7164345
3. ͞2.7835655
4. ͞3.2164345

Explanation:

Here logb x = ͞5.1342618 is given and log10(x(1/4)) is asked

That means here b also treat as base 10

Now, we can say log10 x = ͞5.1342618 = -5 + 0.1342618 => -4.8657382

Therefore, log10 (x1/4) = ¼ log10 x

Or, ¼ (-4.8657382) = -1.21643455, but it is not in the option.

To make it -2, we have to subtract 0.7835655 from -1.21643455

i.e., -1.21643455 - 0.7835655 = -2

Now, to get back on the original value we have to add 0.7835655 in -2.

Or, -2 + 0.7835655 = ͞2.7835655

Hence, the c is correct.

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