# Aptitude Logarithm Test Paper 3

11) The value of Log10 2 + 16 log10 + 12 log10 + 7 log10 is

1. 3
2. 2
3. 1
4. 0

Explanation:
Log10 2 + 16 log10 + 12 log10 + 7 log10

We know that log mn = n log m

So, Log10 2 + log10 [24/ 3*5] 16 + log10 [52/ 23 * 3] 12 + log10 [34/ 24 * 5] 7

Or, log10 2 + log10 [264/ 316 * 516] + log10 [536/ 236 * 312] + log10 [328/ 228 * 57]

We know that log m*n = log m + log n

Now, log10 [2 * 264 * 524 * 328]/ [316 * 516 * 236 * 312 * 228 * 57]

Or, log10 [265 * 328 *524]/ [264 * 328 *523]

Therefore, log10 (2*5) = log10 (10) = 1

12) If log (11 + 4√7) = log (2 + x), what is the value of x?

1. √7
2. 11
3. 4
4. 2

Explanation:

We have log (11 + 4√7) = log (2 + x)

Now, we can write it as log (7 + 4 + 4√7) = log (2 + x)

We know that (a+b) 2 = a2 + b2 + 2ab

Similarly, log (2 + √7) 2 = log(2 + x)

Or, log (2 +√7) = log (2 + x)

Both side Log will be canceled out

Now, 2 + √7 = 2 + x

Therefore, x = 2 + √7 - 2 = √7

13) If log10 = 2, find the value of x.

1. 2, 10
2. 1, 5
3. 12, 24
4. 6, 30

Explanation:

We have log10 = 2

Or, we can write it as log10 = 2 log10 10

Or, log10 = 2 log10 100

Therefore, both side log will be canceled out.

Or, = 4

We know that √x = a, or x1/2 = a, then x = a2

Similarly, x2 - 12x + 36 = 42

Or, x2 - 12x + 36 = 16

Or, x2 - 12x + 20=0

Now, (x-2)(x-10) = 0

Or, x-2 = 0, and x-10 =0

Therefore, we can say that x = 2, x = 10

14) If f(a) = log , f (2a / 1+a2) is equals to

1. 0
2. 1
3. 2f(a)
4. None of these

Explanation:
We have an expression f(a) = log , then we have to find the value of f(2a / 1+a2).

We can find it by replacing a with

Now, f= log

Or, log

Now, the denominator of both fractions will cancel out each other.
Now, f= log

Or, log = log

We know that log mn = n log m.
Therefore, 2 log is equals to 2 f(a).

15) The value of log10is

1. 4
2. 3
3. 2
4. 1

Explanation:
We know that, if a number is in the form

Then the output will be the number itself, i.e., the expressionis equals to 10.

Since, logm m = 1

Therefore, log10 (10) = 1

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