Aptitude Logarithm Test Paper 26) Add the equation x = 1 + log_{a} bc, y = 1 + log_{b} ca and z = 1 + log_{c} ab. Find the value of .
Answer: A Explanation: x =1 + log_{a} bc We can write it as x = log_{a} a + log_{a} bc => log_{a} abc Similarly, y = log_{b} b + log_{b} ca => log_{b} abc And, z = log_{c} c + log_{c} ab => log_{c} abc Y = Z = Now, =[log_{m} a + log_{m} b + log_{m} c] Now, Right hand side equation, [log_{m} abc/ log_{m} abc] = 1 Or, = 1 => xyz =xy + yz + zx. 7) Find the characteristics of the logarithms of i) 5631, ii) 5.678, iii) 56.23.
Answer: B Explanation: i) The number 5631> 1 and the number of digits in the integral part are 4. ii) In the number 5.678, the number of digits in the integral part = 1. iii) In this case, the number of digits in an integral part of the number 56.23 are 2. So, the option b is correct. 8) If log 2 = 0.30103, log 3 = 0.47712, the number of digits in 6^{20} is
Answer: C Explanation: We have to take 6^{20} with log 9) The logarithm of 0.0001 to the base 0.001 is equal to
Answer: A Explanation: Let Log_{0.001} (0.0001) = x We know that log_{x} y = a equals to x^{a} = y. So, (0.001)^{x}= 0.0001 Or, = Therefore, compare both side, we get 3x = 4. Hence, x = 4/3 10) If log_{b} x = ͞5.1342618, then the value of log_{10}(x^{(1/4)}) will be
Answer: C Explanation: Here log_{b} x = ͞5.1342618 is given and log_{10}(x^{(1/4)}) is asked That means here b also treat as base 10 Now, we can say log_{10} x = ͞5.1342618 = 5 + 0.1342618 => 4.8657382 Therefore, log_{10} (x^{1/4}) = ¼ log_{10} 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. Aptitude Logarithm Test Paper 1 Aptitude Logarithm Test Paper 3 Aptitude Logarithm Test Paper 4
Next TopicAptitude Logarithm Test Paper 3

Ansible
Mockito
Talend
Azure
SharePoint
Powershell
Kali Linux
OpenCV
Kafka
Pandas
Joomla
Reinforcement