Aptitude Logarithm Test Paper 4
16) If log10 x = a, log10 y = b, and log10 z = c, so (to base 10) the antilogarithm of pa+qb+rc would be
Let x = antilog (pa+qb+rc)z
Or, we can say that log x = pa+qb+rcz
Or, log x = p log10 x + q log10 y + r log10 z
We know that, p log q = log qp
Or, log x = log10 xp + log10 yq + log10 zr
Or, we know that log a + log b = log ab
Or, log x = log (xp yq zr)
Since, both sides log will be canceled out
Therefore, x = xp yq zr)
Hence, the antilogarithm of pa+qb+rc = x xp yq zr)
17) The value of logb a2 logc b2 loga c2 is
We have an expression logb a2 logc b2 loga c2
Similarly, (2 logb a) (2 logc b) (2 loga c)
Therefore, 8 * 1 = 8
18) If log(x2 - 4x + 5) = 0, what will be the value of x?
We have log(x2 - 4x + 5) = 0
Note: when log moves from one side to the other, it will become exponential.
So, (x2 - 4x + 5) = e0
Or, we know that exponential to the power zero is one.
So, x2 - 4x + 5 = 1
Now, x2 - 4x + 5-1 = 0
Or, x2 - 4x + 4 = 0
Or, (x-2) 2 = 0
Therefore, x- 2 = 0
Hence, x = 2
19) If log2 [log3 (log2 x)] = 1, what is the value of x?
We have an expression log2 [log3 (log2 x)] = 1
We know that, logx y = a, is equals to xa = y
Similarly, [log3 (log2 x)] = 21
Now, log2 x = 32 = 9
Now, x = 29 = 512
Therefore, the value of x from the expression log2 [log3 (log2 x)] = 1, is 512.
20) is equal to
We know that log m + log n + log p = log (m * n * p)
Now, we can write it as, log => log (1)
Note: logx (1) is always equals to be zero.
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