Question paper, Set 1

$3^{4} = 81$

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

$\therefore \log_{3} 81 = 4$

Answer: $4$

The base cannot be $1$, so $\log_{1} 6$ is not defined.

$6^0 = 1 \Rightarrow 0 = \log_{6} 1$

$\therefore \log_{6} 1 = 0$

Answer: $\log_{6} 1 = 0$ ($\log_{1} 6$ is not defined)

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

Here $a = 2$, $x = 10$, $N = 1024$.

$\therefore 10 = \log_{2} 1024$

Answer: $10 = \log_{2} 1024$

$\log_{9} 9 = 1,\ \log_{9} 1 = 0$

$= 4(1) + 7(0) = 4$

Answer: $4$

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

$= \log_{6}(4 \times 9) = \log_{6} 36$

$6^{2} = 36 \Rightarrow \log_{6} 36 = 2$

Answer: $2$

$\log_{8} 32 = \log_{2^{3}} 2^{5}$

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

$\displaystyle = \frac{5}{3}(1) = \frac{5}{3}$

Answer: $\displaystyle \frac{5}{3}$

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

$\therefore \log_{11} 4 \times \log_{4} 11 = 1$

Answer: $1$

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

$= \log_{3} 9$

$= 2$

Answer: $2$

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

$= \log_{5} 3125$

$= 5$

Answer: $\log_{5} 3125 = 5$

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

$= 8 + 5 = 13$

Answer: $13$

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

$3^{\log_{5} 7} = 7^{\log_{5} 3}$

$= 7^{\log_{5} 3} - 7^{\log_{5} 3} = 0$

Answer: $0$

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

$\Rightarrow 3x - 2 = 13$

$\Rightarrow 3x = 15$

$\Rightarrow x = 5$

Answer: $x = 5$

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

$\Rightarrow x^2 - 4x + 6 = 3$

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

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

$\Rightarrow x(x - 3) - 1(x - 3) = 0$

$\Rightarrow (x - 1)(x - 3) = 0$

$\Rightarrow x = 1$ (or) $x = 3$

Answer: $x = 1$ (or) $x = 3$

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

$\Rightarrow \log_{6} x(x + 5) = 2$

$\Rightarrow \log_{6}(x^2 + 5x) = 2$

$\Rightarrow x^2 + 5x = 6^{2} = 36$

$\Rightarrow x^2 + 5x - 36 = 0$

$\Rightarrow x^2 + 9x - 4x - 36 = 0$

$\Rightarrow x(x + 9) - 4(x + 9) = 0$

$\Rightarrow (x - 4)(x + 9) = 0$

$\Rightarrow x = 4$ (or) $x = -9$

If $x = -9$, $\log_{6} x$ is not defined.

$\therefore x \ne -9$

$\therefore x = 4$

Answer: $x = 4$

$= \log_{2^{3}} 2^{4} - \log_{2^{4}} 2^{3}$

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

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

$\displaystyle = \frac{7}{12}$

Answer: $\displaystyle \frac{7}{12}$

$\displaystyle \sqrt[n]{a} = a^{\frac{1}{n}}$

$\displaystyle = \log_{b^{\frac{1}{3}}} a \cdot \log_{c^{\frac{1}{5}}} b \cdot \log_{d^{\frac{1}{2}}} c$

$= 3\log_{b} a \cdot 5\log_{c} b \cdot 2\log_{d} c$

$= 3 \times 5 \times 2 \cdot \log_{b} a \log_{c} b \log_{d} c$

$= 30\log_{d} a$

$\therefore k = 30$

Answer: $k = 30$

$a^{m+n} = a^m \cdot a^n$

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

$= 9 \times 5 = 45$

Answer: $45$

Given $x^2 + y^2 = 7xy$

$LHS = 2\log(x + y) = \log(x + y)^2$

$= \log(x^2 + y^2 + 2xy)$

$= \log(7xy + 2xy) = \log(9xy)$

$= \log 9 + \log x + \log y$

$= \log 3^{2} + \log x + \log y = 2\log 3 + \log x + \log y$

$= RHS$

Answer: Shown: $2\log(x + y) = \log x + \log y + 2\log 3$

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

$\displaystyle \log_{2}(6x - 1) = \log_{2^{-1}} x = \frac{1}{-1}\log_{2} x = \log_{2} x^{-1}$

$\displaystyle \Rightarrow 6x - 1 = x^{-1} = \frac{1}{x}$

$\Rightarrow 6x^2 - x = 1$

$\Rightarrow 6x^2 - x - 1 = 0$

$\Rightarrow 6x^2 + 2x - 3x - 1 = 0$

$\Rightarrow 2x(3x + 1) - 1(3x + 1) = 0$

$\Rightarrow (2x - 1)(3x + 1) = 0$

$\displaystyle \Rightarrow x = \frac{1}{2}$ (or) $\displaystyle x = -\frac{1}{3}$

For $\displaystyle x = -\frac{1}{3}$, $\displaystyle \log_{\frac{1}{2}} x$ is not defined.

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

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

$y = \log_{2} x$

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

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

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

$x = 4:\ y = \log_{2} 4 = 2$

$x = 16:\ y = \log_{2} 16 = 4$

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