Aptitude Logarithm Test Paper 416) If log_{10} x = a, log_{10} y = b, and log_{10} z = c, so (to base 10) the antilogarithm of pa+qb+rc would be
Answer: B Explanation: Let x = antilog (pa+qb+rc)z Or, we can say that log x = pa+qb+rcz Or, log x = p log_{10} x + q log_{10} y + r log_{10} z We know that, p log q = log q^{p} Or, log x = log_{10} xp + log_{10} yq + log_{10} zr Or, we know that log a + log b = log ab Or, log x = log (x^{p} y^{q} z^{r}) Since, both sides log will be canceled out Therefore, x = x^{p} y^{q} z^{r}) Hence, the antilogarithm of pa+qb+rc = x x^{p} y^{q} z^{r}) 17) The value of log_{b} a^{2} log_{c} b^{2} log_{a} c^{2} is
Answer: C Explanation: We have an expression log_{b} a^{2} log_{c} b^{2} log_{a} c^{2} Similarly, (2 log_{b} a) (2 log_{c} b) (2 log_{a} c) Therefore, 8 * 1 = 8 18) If log(x^{2}  4x + 5) = 0, what will be the value of x?
Answer: A Explanation: We have log(x^{2}  4x + 5) = 0 Note: when log moves from one side to the other, it will become exponential. So, (x^{2}  4x + 5) = e^{0} Or, we know that exponential to the power zero is one. So, x^{2}  4x + 5 = 1 Now, x^{2}  4x + 51 = 0 Or, x^{2}  4x + 4 = 0 Or, (x2) ^{2} = 0 Therefore, x 2 = 0 Hence, x = 2 19) If log_{2} [log_{3} (log_{2} x)] = 1, what is the value of x?
Answer: A Explanation: We have an expression log_{2} [log_{3} (log_{2} x)] = 1 We know that, logx y = a, is equals to x^{a} = y Similarly, [log_{3} (log_{2} x)] = 2^{1} Now, log_{2} x = 3^{2} = 9 Now, x = 2^{9} = 512 Therefore, the value of x from the expression log_{2} [log_{3} (log_{2} x)] = 1, is 512. 20) is equal to
Answer: C Explanation: We know that log m + log n + log p = log (m * n * p) Now, we can write it as, log => log (1) Note: log_{x} (1) is always equals to be zero. Aptitude Logarithm Test Paper 1 Aptitude Logarithm Test Paper 2 Aptitude Logarithm Test Paper 3
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