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 .
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
Now, =[logm a + logm b + logm c]
Now, Right hand side equation, [logm abc/ logm abc] = 1
Or, = 1 => xyz =xy + yz + zx.
7) Find the characteristics of the logarithms of i) 5631, ii) 5.678, iii) 56.23.
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 620 is
We have to take 620 with log
9) The logarithm of 0.0001 to the base 0.001 is equal to
Let Log0.001 (0.0001) = x
We know that logx y = a equals to xa = y.
So, (0.001)x= 0.0001
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
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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