Question paper, Set 5

$\displaystyle \frac{1}{100} = \frac{1}{10^{2}} = 10^{-2}$

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

$\displaystyle \therefore \log_{10} \frac{1}{100} = -2$

Answer: $-2$

The number must be positive ($N > 0$), so $\log_{5}(-25)$ is not defined.

$5^2 = 25 \Rightarrow 2 = \log_{5} 25$

$\therefore \log_{5} 25 = 2$

Answer: $\log_{5} 25 = 2$ ($\log_{5}(-25)$ is not defined)

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

Here $a = 16$, $\displaystyle x = \frac{3}{4}$, $N = 8$.

$\displaystyle \therefore \frac{3}{4} = \log_{16} 8$

Answer: $\displaystyle \frac{3}{4} = \log_{16} 8$

$\log_{0.7} 0.7 = 1,\ \log_{12} 12 = 1$

$= 1 \times 1 = 1$

Answer: $1$

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

$\displaystyle = \log_{5} \frac{250}{2} = \log_{5} 125$

$5^{3} = 125 \Rightarrow \log_{5} 125 = 3$

Answer: $3$

$\log_{625} 125 = \log_{5^{4}} 5^{3}$

$\displaystyle = \frac{3}{4} \log_{5} 5$

$\displaystyle = \frac{3}{4}(1) = \frac{3}{4}$

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

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

$\therefore \log_{9} 20 \times \log_{20} 9 = 1$

Answer: $1$

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

$= \log_{3} 27$

$= 3$

Answer: $3$

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

$= \log_{49} 2401$

$= 2$

Answer: $\log_{49} 2401 = 2$

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

$= 7 \times 3 = 21$

Answer: $21$

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

$\therefore 8^{\log_{3} 7} = 7^{\log_{3} 8}$

Answer: $7^{\log_{3} 8}$

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

$\Rightarrow x = 12 - x$

$\Rightarrow 2x = 12$

$\Rightarrow x = 6$

Answer: $x = 6$

$\log_{a} N = 0 \Rightarrow N = 1$

$\Rightarrow \log_{5} \log_{3} x = 1$

$\Rightarrow \log_{3} x = 5^1 = 5$

$\Rightarrow x = 3^5 = 243$

Answer: $x = 243$

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

$\displaystyle \Rightarrow \log_{2} \frac{x + 6}{x - 1} = 3$

$\displaystyle \Rightarrow \frac{x + 6}{x - 1} = 2^{3} = 8$

$\Rightarrow x + 6 = 8x - 8$

$\Rightarrow 7x = 14$

$\Rightarrow x = 2$

For $x = 2$ both logarithms are defined.

Answer: $x = 2$

$\log_{4} 64^3 = \log_{4}(4^3)^3 = \log_{4} 4^9 = 9$

$\log_{3} 9 = \log_{3} 3^2 = 2$

$\log_{2} 2 = 1$

$\therefore \log_{2} \log_{3} \log_{4} 64^3 = 1$

Answer: $1$

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

$\displaystyle = \log_{b^{\frac{1}{4}}} a \cdot \log_{c^{\frac{1}{3}}} b \cdot \log_{d^{\frac{1}{5}}} c$

$= 4\log_{b} a \cdot 3\log_{c} b \cdot 5\log_{d} c$

$= 4 \times 3 \times 5 \cdot \log_{b} a \log_{c} b \log_{d} c$

$= 60\log_{d} a$

$\therefore k = 60$

Answer: $k = 60$

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

$3^{\log_{5} 7} = 7^{\log_{5} 3},\ 2^{\log_{3} 9} = 9^{\log_{3} 2}$

$= 7^{\log_{5} 3} + 9^{\log_{3} 2} - 9^{\log_{3} 2} - 7^{\log_{5} 3} = 0$

Answer: $0$

Given $x^2 + y^2 = 11xy$

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

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

$= \log(11xy - 2xy) = \log(9xy)$

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

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

$= RHS$

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

$\displaystyle \frac{1}{5} = 5^{-1}$

$\displaystyle \log_{5}(4x + 3) = \log_{5^{-1}} x = \frac{1}{-1}\log_{5} x = \log_{5} x^{-1}$

$\displaystyle \Rightarrow 4x + 3 = x^{-1} = \frac{1}{x}$

$\Rightarrow 4x^2 + 3x = 1$

$\Rightarrow 4x^2 + 3x - 1 = 0$

$\Rightarrow 4x^2 - x + 4x - 1 = 0$

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

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

$\Rightarrow x = -1$ (or) $\displaystyle x = \frac{1}{4}$

For $x = -1$, $\displaystyle \log_{\frac{1}{5}} x$ is not defined.

$\displaystyle \therefore x = \frac{1}{4}$

Answer: $\displaystyle x = \frac{1}{4}$

$y = \log_{4} x$

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

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

$x = 1:\ y = \log_{4} 1 = 0$

$x = 4:\ y = \log_{4} 4 = 1$

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

$x = 64:\ y = \log_{4} 64 = 3$

The base is more than $1$, so the graph rises from left to right.

Answer: $y = -2, -1, 0, 1, 2, 3$; the graph rises