Question paper, Set 3

$5^{3} = 125$

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

$\therefore \log_{5} 125 = 3$

Answer: $3$

The base cannot be negative, so $\log_{-3} 9$ is not defined.

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

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

Answer: $\log_{3} 9 = 2$ ($\log_{-3} 9$ is not defined)

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

Here $a = 5$, $x = -3$, $\displaystyle N = \frac{1}{125}$.

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

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

$\log_{a} a = 1,\ \log_{a} 1 = 0$ with $\displaystyle a = \frac{3}{5}$

$= 1 + 0 = 1$

Answer: $1$

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

$\displaystyle = \log_{10} \frac{8 \times 25}{2} = \log_{10} 100$

$10^{2} = 100 \Rightarrow \log_{10} 100 = 2$

Answer: $2$

$\log_{9} 243 = \log_{3^{2}} 3^{5}$

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

$\displaystyle = \frac{5}{2}(1) = \frac{5}{2}$

Answer: $\displaystyle \frac{5}{2}$

$\displaystyle \frac{1}{\log_{32} 2} = \log_{2} 32$

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

Answer: $5$

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

$= \log_{3} 243$

$= 5$

Answer: $5$

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

$= \log_{343} 49$

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

Answer: $\displaystyle \log_{343} 49 = \frac{2}{3}$

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

$\displaystyle = \frac{2}{3} \times 12 = 8$

Answer: $8$

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

$7^{\log_{2} 3} = 3^{\log_{2} 7}$

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

Answer: $1$

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

$\Rightarrow 2x + 1 = x + 6$

$\Rightarrow x = 5$

$\Rightarrow x = 5$

Answer: $x = 5$

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

$\Rightarrow x^2 - 7x + 13 = 6^{0} = 1$

$\Rightarrow x^2 - 7x + 12 = 0$

$\Rightarrow x^2 - 4x - 3x + 12 = 0$

$\Rightarrow x(x - 4) - 3(x - 4) = 0$

$\Rightarrow (x - 3)(x - 4) = 0$

$\Rightarrow x = 3$ (or) $x = 4$

Answer: $x = 3$ (or) $x = 4$

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

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

$\displaystyle \Rightarrow \frac{x}{x + 2} = 3^{2} = 9$

$\Rightarrow x = 9x + 18$

$\Rightarrow 8x = -18$

$\displaystyle \Rightarrow x = -\frac{9}{4}$

For $\displaystyle x = -\frac{9}{4}$, $\log_{3} x$ is not defined.

$\therefore$ there is no real $x$.

Answer: No real $x$

$\log_{0.2} 0.008 = \log_{0.2}(0.2)^3 = 3\log_{0.2} 0.2 = 3$

$\displaystyle \log_{2} \sqrt{32} = \log_{2}(2^5)^{\frac{1}{2}} = \log_{2} 2^{\frac{5}{2}} = \frac{5}{2}$

$\displaystyle = 3 + \frac{5}{2} = \frac{11}{2}$

Answer: $\displaystyle \frac{11}{2}$

$\displaystyle 6\sqrt{6} = 6^1 \times 6^{\frac{1}{2}} = 6^{\frac{3}{2}}$ (and the same for the others)

$\displaystyle = \frac{\log (\frac{6 \times 5}{4})^{\frac{3}{2}}}{\log 7.5}$

$\displaystyle = \log_{7.5}(7.5)^{\frac{3}{2}}$ since $\displaystyle \frac{6 \times 5}{4} = 7.5$

$\displaystyle = \frac{3}{2}\log_{7.5} 7.5 = \frac{3}{2}$

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

$\displaystyle \log_{25} 64 = \log_{5^2} 8^2 = \frac{2}{2}\log_{5} 8 = \log_{5} 8$

$5^{\log_{25} 64} = 5^{\log_{5} 8}$

$= 8$

Answer: $8$

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

$\displaystyle \Rightarrow \log(3a - 2b) = \log \frac{a}{b}$

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

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

$\displaystyle \Rightarrow a\left(3 - \frac{1}{b}\right) = 2b$

$\displaystyle \Rightarrow a\left(\frac{3b - 1}{b}\right) = 2b$

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

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

$\displaystyle \frac{1}{3} = 3^{-1}$

$\displaystyle \log_{3}(3x + 2) = \log_{3^{-1}} x = \frac{1}{-1}\log_{3} x = \log_{3} x^{-1}$

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

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

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

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

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

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

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

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

$\displaystyle \therefore x = \frac{1}{3}$

Answer: $\displaystyle x = \frac{1}{3}$

$y = \log_{3} x$

$\displaystyle x = \frac{1}{9}:\ y = \log_{3} \frac{1}{9} = -2$

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

$x = 1:\ y = \log_{3} 1 = 0$

$x = 3:\ y = \log_{3} 3 = 1$

$x = 9:\ y = \log_{3} 9 = 2$

$x = 27:\ y = \log_{3} 27 = 3$

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