Question paper, Set 6

$2^{7} = 128$

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

$\therefore \log_{2} 128 = 7$

Answer: $7$

$0^2 = 0 \Rightarrow 2 = \log_{0} 0$

$0^5 = 0 \Rightarrow 5 = \log_{0} 0$

So $2 = 5$, which has no meaning.

$\therefore a \ne 0$

Answer: $0^2 = 0$ and $0^5 = 0$ would give $2 = 5$, which has no meaning

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

Here $a = 11$, $N = 1331$, $x = 3$.

$\therefore 1331 = 11^{3}$

Answer: $1331 = 11^{3}$

$\log_{20} 20 = 1,\ \log_{3} 1 = 0$

$\displaystyle = \frac{1}{5} + 0 = \frac{1}{5}$

Answer: $\displaystyle \frac{1}{5}$

$\displaystyle \log_{a} m + \log_{a} n - \log_{a} p = \log_{a} \frac{m \times n}{p}$

$\displaystyle = \log_{2} \frac{12 \times 10}{15} = \log_{2} 8$

$2^{3} = 8 \Rightarrow \log_{2} 8 = 3$

Answer: $3$

$\log_{8} 11^{4} = \log_{2^{3}} 11^{4}$

$\displaystyle = \frac{4}{3} \log_{2} 11$

Answer: $\displaystyle \frac{4}{3}\log_{2} 11$

$\displaystyle \frac{1}{\log_{36} 6} = \log_{6} 36$

$= \log_{6} 6^{2} = 2\log_{6} 6 = 2$

Answer: $2$

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

$= \log_{4} 64$

$= 3$

Answer: $3$

$\displaystyle \log_{b} a = \frac{\log_{x} a}{\log_{x} b}$ with $x = 10$

$\displaystyle \therefore \log_{3} 20 = \frac{\log_{10} 20}{\log_{10} 3}$

Answer: $\displaystyle \frac{\log_{10} 20}{\log_{10} 3}$

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

$= 14 - 9 = 5$

Answer: $5$

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

$10^{\log_{3} 4} = 4^{\log_{3} 10}$

$\therefore$ the numerator and denominator are equal, and the value is $1$.

Answer: $1$

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

$\Rightarrow x^2 = 5x - 6$

$\Rightarrow x^2 - 5x + 6 = 0$

$\Rightarrow x^2 - 3x - 2x + 6 = 0$

$\Rightarrow x(x - 3) - 2(x - 3) = 0$

$\Rightarrow (x - 2)(x - 3) = 0$

$\Rightarrow x = 2$ (or) $x = 3$

Answer: $x = 2$ (or) $x = 3$

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

$\Rightarrow x^2 + x - 5 = 9^{0} = 1$

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

$\Rightarrow x^2 + 3x - 2x - 6 = 0$

$\Rightarrow x(x + 3) - 2(x + 3) = 0$

$\Rightarrow (x - 2)(x + 3) = 0$

$\Rightarrow x = 2$ (or) $x = -3$

Answer: $x = 2$ (or) $x = -3$

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

$\Rightarrow \log_{10}(x + 15)x = 2$

$\Rightarrow \log_{10}(x^2 + 15x) = 2$

$\Rightarrow x^2 + 15x = 10^{2} = 100$

$\Rightarrow x^2 + 15x - 100 = 0$

$\Rightarrow x^2 + 20x - 5x - 100 = 0$

$\Rightarrow x(x + 20) - 5(x + 20) = 0$

$\Rightarrow (x - 5)(x + 20) = 0$

$\Rightarrow x = 5$ (or) $x = -20$

If $x = -20$, $\log_{10} x$ is not defined.

$\therefore x \ne -20$

$\therefore x = 5$

Answer: $x = 5$

$\log_{0.5} 0.0625 = \log_{0.5}(0.5)^4 = 4$

$\displaystyle \log_{3} \sqrt[3]{243} = \log_{3}(3^5)^{\frac{1}{3}} = \log_{3} 3^{\frac{5}{3}} = \frac{5}{3}$

$\displaystyle = 4 - \frac{5}{3} = \frac{7}{3}$

Answer: $\displaystyle \frac{7}{3}$

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

$LHS = \log_{xyz} xy + \log_{xyz} yz + \log_{xyz} zx$

$= \log_{xyz}(xy \cdot yz \cdot zx)$

$= \log_{xyz}(xyz)^2$

$= 2\log_{xyz} xyz = 2(1) = 2 = RHS$

Answer: $2$

$\displaystyle \log_{7} 1 = 0 \Rightarrow 4^{-2 + 0} = 4^{-2} = \frac{1}{16}$

$3^{1 + \log_{3} 2} = 3^1 \times 3^{\log_{3} 2} = 3 \times 2 = 6$

$\displaystyle = \frac{1}{16} + 6 = \frac{97}{16}$

Answer: $\displaystyle \frac{97}{16}$

$\displaystyle \log \frac{a - b}{4} = \frac{1}{2}\log ab = \log \sqrt{ab}$

$\displaystyle \Rightarrow \frac{a - b}{4} = \sqrt{ab}$

$\Rightarrow a - b = 4\sqrt{ab}$

Squaring on both sides,

$\Rightarrow a^2 - 2ab + b^2 = 16ab$

$\Rightarrow a^2 + b^2 = 18ab$

$\displaystyle \Rightarrow \frac{a^2}{ab} + \frac{b^2}{ab} = 18$

$\displaystyle \Rightarrow \frac{a}{b} + \frac{b}{a} = 18$

Answer: $18$

$\displaystyle \log_{49}(x + 10) = \log_{7^2}(x + 10) = \frac{1}{2}\log_{7}(x + 10)$

$\Rightarrow 2\log_{7}(x - 2) = \log_{7}(x + 10)$

$\Rightarrow \log_{7}(x - 2)^2 = \log_{7}(x + 10)$

$\Rightarrow (x - 2)^2 = x + 10$

$\Rightarrow x^2 - 4x + 4 = x + 10$

$\Rightarrow x^2 - 5x - 6 = 0$

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

$\Rightarrow x(x - 6) + 1(x - 6) = 0$

$\Rightarrow (x + 1)(x - 6) = 0$

$\Rightarrow x = -1$ (or) $x = 6$

For $x = -1$, $\log_{7}(x - 2)$ is not defined.

$\therefore x = 6$

Answer: $x = 6$

$\displaystyle y = \log_{\frac{1}{5}} x$

$\displaystyle x = \frac{1}{25}:\ y = \log_{\frac{1}{5}} \frac{1}{25} = 2$

$\displaystyle x = \frac{1}{5}:\ y = \log_{\frac{1}{5}} \frac{1}{5} = 1$

$\displaystyle x = 1:\ y = \log_{\frac{1}{5}} 1 = 0$

$\displaystyle x = 5:\ y = \log_{\frac{1}{5}} 5 = -1$

$\displaystyle x = 125:\ y = \log_{\frac{1}{5}} 125 = -3$

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