Question paper, Set 4

$11^{2} = 121$

$a^x = N \Rightarrow x = \log_{a} N$

$\therefore \log_{11} 121 = 2$

Answer: $2$

$1^4 = 1 \Rightarrow 4 = \log_{1} 1$

$1^7 = 1 \Rightarrow 7 = \log_{1} 1$

So $4 = 7$, which has no meaning.

$\therefore a \ne 1$

Answer: $1^4 = 1$ and $1^7 = 1$ would give $4 = 7$, which has no meaning

$\log_{a} N = x \Leftrightarrow N = a^x$

Here $a = 8$, $\displaystyle N = \frac{1}{64}$, $x = -2$.

$\displaystyle \therefore \frac{1}{64} = 8^{-2}$

Answer: $\displaystyle \frac{1}{64} = 8^{-2}$

$\log_{13} 13 = 1,\ \log_{8} 8 = 1$

$= 6(1) - 2(1) = 4$

Answer: $4$

$\log_{a} m + \log_{a} n = \log_{a} mn$

$= \log_{12}(8 \times 18) = \log_{12} 144$

$12^{2} = 144 \Rightarrow \log_{12} 144 = 2$

Answer: $2$

$\displaystyle \frac{1}{3} = 3^{-1},\ 81 = 3^{4}$

$\displaystyle \log_{\frac{1}{3}} 81 = \log_{3^{-1}} 3^{4}$

$\displaystyle = \frac{4}{-1} \log_{3} 3 = -4$

Answer: $-4$

$\displaystyle \log_{b} a = \frac{1}{\log_{a} b}$

$\displaystyle \log_{10} 3 = \frac{1}{\log_{3} 10} = \frac{1}{p}$

Answer: $\displaystyle \frac{1}{p}$

$\log_{b} a \cdot \log_{c} b = \log_{c} a$

$= \log_{16} 8$

$\displaystyle = \frac{3}{4}$

Answer: $\displaystyle \frac{3}{4}$

$\displaystyle \frac{\log a}{\log b} = \log_{b} a$

$= \log_{16} 64$

$\displaystyle = \frac{3}{2}$

Answer: $\displaystyle \log_{16} 64 = \frac{3}{2}$

$a^{\log_{a} N} = N$

$\displaystyle = \frac{45}{9} = 5$

Answer: $5$

$a^{\log_{c} b} = b^{\log_{c} a}$

$6^{\log_{5} 11} = 11^{\log_{5} 6}$

$= 11^{\log_{5} 6} - 11^{\log_{5} 6} = 0$

Answer: $0$

$\log_{a} m = \log_{a} n \Rightarrow m = n$

$\Rightarrow \sqrt[3]{x} = 4$

$\Rightarrow x = 4^{3} = 64$

Answer: $x = 64$

$\log_{a} N = x \Rightarrow N = a^x$

$\Rightarrow x^2 + 6x + 13 = 5$

$\Rightarrow x^2 + 6x + 8 = 0$

$\Rightarrow x^2 + 4x + 2x + 8 = 0$

$\Rightarrow x(x + 4) + 2(x + 4) = 0$

$\Rightarrow (x + 2)(x + 4) = 0$

$\Rightarrow x = -2$ (or) $x = -4$

Answer: $x = -2$ (or) $x = -4$

$\log_{a} m + \log_{a} n = \log_{a} mn$

$\Rightarrow \log_{3}(x - 6)x = 3$

$\Rightarrow \log_{3}(x^2 - 6x) = 3$

$\Rightarrow x^2 - 6x = 3^{3} = 27$

$\Rightarrow x^2 - 6x - 27 = 0$

$\Rightarrow x^2 - 9x + 3x - 27 = 0$

$\Rightarrow x(x - 9) + 3(x - 9) = 0$

$\Rightarrow (x + 3)(x - 9) = 0$

$\Rightarrow x = -3$ (or) $x = 9$

If $x = -3$, $\log_{3} x$ is not defined.

$\therefore x \ne -3$

$\therefore x = 9$

Answer: $x = 9$

$= \log_{2^{2}} 2^{5} - \log_{2^{5}} 2^{2}$

$\displaystyle = \frac{5}{2} \log_{2} 2 - \frac{2}{5} \log_{2} 2$

$\displaystyle = \frac{5}{2} - \frac{2}{5}$

$\displaystyle = \frac{21}{10}$

Answer: $\displaystyle \frac{21}{10}$

$\displaystyle \log_{b^n} a^m = \frac{m}{n}\log_{b} a$

$\displaystyle LHS = \frac{1}{2}\log_{b} a \cdot \frac{1}{3}\log_{c} b \cdot \frac{1}{4}\log_{a} c$

$\displaystyle = \frac{1}{24} \log_{b} a \log_{c} b \log_{a} c$

$\log_{b} a \log_{c} b \log_{a} c = \log_{a} a = 1$

$\displaystyle = \frac{1}{24}(1) = \frac{1}{24} = RHS$

Answer: $\displaystyle \frac{1}{24}$

$2^{4 + \log_{2} 3} = 2^4 \times 2^{\log_{2} 3} = 16 \times 3 = 48$

$9^{\log_{3} 5} = (3^2)^{\log_{3} 5} = (3^{\log_{3} 5})^2 = 5^2 = 25$

$= 48 - 25 = 23$

Answer: $23$

$\log_{a} m + \log_{a} n = \log_{a} mn$

$= \log_{3}[(c - b)(c^2 + cb + b^2)]$

$(a - b)(a^2 + ab + b^2) = a^3 - b^3$

$= \log_{3}(c^3 - b^3) = \log_{3} 27$

$= \log_{3} 3^{3} = 3$

Answer: $3$

$\displaystyle \log_{9}(x + 5) = \log_{3^2}(x + 5) = \frac{1}{2}\log_{3}(x + 5)$

$\Rightarrow 2\log_{3}(x - 1) = \log_{3}(x + 5)$

$\Rightarrow \log_{3}(x - 1)^2 = \log_{3}(x + 5)$

$\Rightarrow (x - 1)^2 = x + 5$

$\Rightarrow x^2 - 2x + 1 = x + 5$

$\Rightarrow x^2 - 3x - 4 = 0$

$\Rightarrow x^2 - 4x + x - 4 = 0$

$\Rightarrow x(x - 4) + 1(x - 4) = 0$

$\Rightarrow (x + 1)(x - 4) = 0$

$\Rightarrow x = -1$ (or) $x = 4$

For $x = -1$, $\log_{3}(x - 1)$ is not defined.

$\therefore x = 4$

Answer: $x = 4$

$\displaystyle y = \log_{\frac{1}{2}} x$

$\displaystyle x = \frac{1}{8}:\ y = \log_{\frac{1}{2}} \frac{1}{8} = 3$

$\displaystyle x = \frac{1}{4}:\ y = \log_{\frac{1}{2}} \frac{1}{4} = 2$

$\displaystyle x = 1:\ y = \log_{\frac{1}{2}} 1 = 0$

$\displaystyle x = 2:\ y = \log_{\frac{1}{2}} 2 = -1$

$\displaystyle x = 8:\ y = \log_{\frac{1}{2}} 8 = -3$

$\displaystyle x = 16:\ y = \log_{\frac{1}{2}} 16 = -4$

Answer: $y = 3, 2, 0, -1, -3, -4$