Question paper, Set 7

$\displaystyle \frac{1}{36} = \frac{1}{6^{2}} = 6^{-2}$

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

$\displaystyle \therefore \log_{6} \frac{1}{36} = -2$

Answer: $-2$

The base cannot be $0$, so $\log_{0} 0$ is not defined.

$8^1 = 8 \Rightarrow 1 = \log_{8} 8$

$\therefore \log_{8} 8 = 1$

Answer: $\log_{8} 8 = 1$ ($\log_{0} 0$ is not defined)

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

Here $a = 9$, $\displaystyle x = \frac{1}{2}$, $N = 3$.

$\displaystyle \therefore \frac{1}{2} = \log_{9} 3$

Answer: $\displaystyle \frac{1}{2} = \log_{9} 3$

$\log_{7} 7 = 1,\ \log_{25} 1 = 0$

$= 1^2 + 3(0) = 1$

Answer: $1$

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

$\displaystyle = \log_{3}(54 \times \frac{1}{2}) = \log_{3} 27$

$3^{3} = 27 \Rightarrow \log_{3} 27 = 3$

Answer: $3$

$\log_{4} 128 = \log_{2^{2}} 2^{7}$

$\displaystyle = \frac{7}{2} \log_{2} 2$

$\displaystyle = \frac{7}{2}(1) = \frac{7}{2}$

Answer: $\displaystyle \frac{7}{2}$

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

$\displaystyle \therefore \frac{1}{\log_{12} 7} = \log_{7} 12$

Answer: $\log_{7} 12$

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

$= \log_{8} 16$

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

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

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

$= \log_{8} 128$

$\displaystyle = \frac{7}{3}$

Answer: $\displaystyle \log_{8} 128 = \frac{7}{3}$

$13^{\log_{13} 4} = 4$ since $a^{\log_{a} N} = N$

$= 4^2 = 16$

Answer: $16$

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

$2^{\log_{7} 9} = 9^{\log_{7} 2}$

$= 9^{\log_{7} 2} - 9^{\log_{7} 2} = 0$

Answer: $0$

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

$\Rightarrow 4x - 3 = 2x + 9$

$\Rightarrow 2x = 12$

$\Rightarrow x = 6$

Answer: $x = 6$

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

$\Rightarrow x^2 - 2x - 7 = 2^{3} = 8$

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

$\Rightarrow x^2 - 5x + 3x - 15 = 0$

$\Rightarrow x(x - 5) + 3(x - 5) = 0$

$\Rightarrow (x + 3)(x - 5) = 0$

$\Rightarrow x = -3$ (or) $x = 5$

Answer: $x = -3$ (or) $x = 5$

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

$\Rightarrow \log_{2}[(x^2 + 2x + 4)(x - 2)] = 4$

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

$\Rightarrow \log_{2}(x^3 - 8) = 4$

$\Rightarrow x^3 - 8 = 2^{4} = 16$

$\Rightarrow x^3 = 24$

$\Rightarrow x = \sqrt[3]{24} = 2\sqrt[3]{3}$

Answer: $x = 2\sqrt[3]{3}$

$= \log_{5^{2}} 5^{3} + \log_{5^{3}} 5^{2}$

$\displaystyle = \frac{3}{2} \log_{5} 5 + \frac{2}{3} \log_{5} 5$

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

$\displaystyle = \frac{13}{6}$

Answer: $\displaystyle \frac{13}{6}$

$\displaystyle 2\sqrt[3]{2} = 2^1 \times 2^{\frac{1}{3}} = 2^{\frac{4}{3}}$ (and the same for the others)

$\displaystyle = \frac{\log (2 \times 5)^{\frac{4}{3}}}{\log 10}$

$\displaystyle = \log_{10}(10)^{\frac{4}{3}}$ since $2 \times 5 = 10$

$\displaystyle = \frac{4}{3}\log_{10} 10 = \frac{4}{3}$

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

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

$5^{\log_{25} 16} = 5^{\log_{5^2} 4^2} = 5^{\log_{5} 4} = 4$

$\displaystyle = \frac{4}{3} \times \frac{3}{2} \times 4 = 8$

Answer: $8$

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

$\displaystyle \Rightarrow \log(a + 3b) = \log \frac{a}{b}$

$\displaystyle \Rightarrow a + 3b = \frac{a}{b}$

$\displaystyle \Rightarrow a - \frac{a}{b} = -3b$

$\displaystyle \Rightarrow a\left(1 - \frac{1}{b}\right) = -3b$

$\displaystyle \Rightarrow a\left(\frac{b - 1}{b}\right) = -3b$

$\displaystyle \Rightarrow a = \frac{-3b^2}{b - 1} = \frac{3b^2}{1 - b}$

Answer: $\displaystyle a = \frac{3b^2}{1 - b}$

$\displaystyle \frac{1}{7} = 7^{-1}$

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

$\displaystyle \Rightarrow 2x + 1 = x^{-1} = \frac{1}{x}$

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

$\Rightarrow 2x^2 + x - 1 = 0$

$\Rightarrow 2x^2 - x + 2x - 1 = 0$

$\Rightarrow x(2x - 1) + 1(2x - 1) = 0$

$\Rightarrow (x + 1)(2x - 1) = 0$

$\Rightarrow x = -1$ (or) $\displaystyle x = \frac{1}{2}$

For $x = -1$, $\displaystyle \log_{\frac{1}{7}} x$ is not defined.

$\displaystyle \therefore x = \frac{1}{2}$

Answer: $\displaystyle x = \frac{1}{2}$

$y = \log_{5} x$

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

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

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

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

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

$x = 125:\ y = \log_{5} 125 = 3$

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