Question paper, Set 8

$13^{2} = 169$

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

$\therefore \log_{13} 169 = 2$

Answer: $2$

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

The base cannot be $0$, $1$ or a negative number: $a > 0$, $a \ne 1$.

$a^x$ is always positive, so $N > 0$.

$\therefore a > 0,\ a \ne 1,\ N > 0$

Answer: $a > 0$, $a \ne 1$ and $N > 0$

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

Here $a = 0.5$, $N = 0.125$, $x = 3$.

$\therefore 0.125 = 0.5^{3}$

Answer: $0.125 = 0.5^{3}$

$\log_{11} 11 = 1,\ \log_{14} 14 = 1,\ \log_{17} 1 = 0$

$= 1 + 1 + 0 = 2$

Answer: $2$

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

$\displaystyle = \log_{7} \frac{98}{2} = \log_{7} 49$

$7^{2} = 49 \Rightarrow \log_{7} 49 = 2$

Answer: $2$

$\displaystyle \frac{1}{5} = 5^{-1},\ 125 = 5^{3}$

$\displaystyle \log_{\frac{1}{5}} 125 = \log_{5^{-1}} 5^{3}$

$\displaystyle = \frac{3}{-1} \log_{5} 5 = -3$

Answer: $-3$

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

$= 3(1) = 3$

Answer: $3$

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

$= \log_{9} 27$

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

Answer: $\displaystyle \frac{3}{2}$

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

$= \log_{25} 625$

$= 2$

Answer: $\log_{25} 625 = 2$

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

$= 0.5 + 2.5 = 3$

Answer: $3$

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

$\therefore 10^{\log_{4} 9} = 9^{\log_{4} 10}$

Answer: $9^{\log_{4} 10}$

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

$\Rightarrow x + 4 = 2x$

$\Rightarrow x = 4$

$\Rightarrow x = 4$

Answer: $x = 4$

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

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

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

$\Rightarrow x - 3 = 2^3 = 8$

$\Rightarrow x = 11$

Answer: $x = 11$

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

$\Rightarrow \log_{4}(x - 3)x = 1$

$\Rightarrow \log_{4}(x^2 - 3x) = 1$

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

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

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

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

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

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

If $x = -1$, $\log_{4} x$ is not defined.

$\therefore x \ne -1$

$\therefore x = 4$

Answer: $x = 4$

$\log_{7} 49^4 = \log_{7}(7^2)^4 = \log_{7} 7^8 = 8$

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

$\log_{3} 3 = 1$

$\therefore \log_{3} \log_{2} \log_{7} 49^4 = 1$

Answer: $1$

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

$= \log_{5} 25 + \log_{5} 125 + \log_{5} 625 + \log_{5} 5$

$= \log_{5}(25 \times 125 \times 625 \times 5)$

$= \log_{5}(5^{2} \times 5^{3} \times 5^{4} \times 5^{1}) = \log_{5} 5^{10}$

$= 10\log_{5} 5 = 10$

Answer: $10$

$\displaystyle 10^{-2 + \log_{10} 3} = 10^{-2} \times 10^{\log_{10} 3} = \frac{1}{100} \times 3 = \frac{3}{100}$

$7^{\log_{49} 25} = 7^{\log_{7^2} 5^2} = 7^{\log_{7} 5} = 5$

$\displaystyle = \frac{3}{100} \times 5 = \frac{3}{20}$

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

Given $x^3 = y^3 + 125 \Rightarrow x^3 - y^3 = 125$

$LHS = \log_{5}(x - y) + \log_{5}(x^2 + xy + y^2)$

$= \log_{5}[(x - y)(x^2 + xy + y^2)]$

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

$= \log_{5}(x^3 - y^3) = \log_{5} 125$

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

Answer: Shown: the sum is $3$

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

$\displaystyle \log_{6}(5x - 4) = \log_{6^{-1}} x = \frac{1}{-1}\log_{6} x = \log_{6} x^{-1}$

$\displaystyle \Rightarrow 5x - 4 = x^{-1} = \frac{1}{x}$

$\Rightarrow 5x^2 - 4x = 1$

$\Rightarrow 5x^2 - 4x - 1 = 0$

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

$\Rightarrow x(5x + 1) - 1(5x + 1) = 0$

$\Rightarrow (x - 1)(5x + 1) = 0$

$\Rightarrow x = 1$ (or) $\displaystyle x = -\frac{1}{5}$

For $\displaystyle x = -\frac{1}{5}$, $\displaystyle \log_{\frac{1}{6}} x$ is not defined.

$\therefore x = 1$

Answer: $x = 1$

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

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

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

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

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

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

$\displaystyle x = 64:\ y = \log_{\frac{1}{4}} 64 = -3$

The base is between $0$ and $1$, so the graph falls from left to right.

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