Question

The cube root of $A(x - d) - B(x - c) + C(x - c)(x - d) = (x - a)$ is.

• A. $x^{2}$
• B. $C = 1$
• C. $\frac{x^{2} - 5}{x^{2} - 3x + 2} = \frac{x^{2} - 5}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2} + C$
• D. $\Rightarrow$

Answer 1

Answer: (D)

$\Rightarrow$

Answer 2

Let $= \log_{abc}abc = 1$

$a = \log_{24}12 = \frac{\log 12}{\log 24} = \frac{2\log 2 + \log 3}{3\log 2 + \log 3}b = \log_{36}24 = \frac{3\log 2 + \log 3}{2(\log 2 + \log 3)}$

$c = \log_{48}36 = \frac{2(\log 2 + \log 3)}{4\log 2 + \log 3}$

$\therefore$

$abc = \frac{2\log 2 + \log 3}{4\log 2 + \log 3}$

$1 + abc = \frac{6\log 2 + 2\log 3}{4\log 2 + \log 3} = 2.\frac{3\log 2 + \log 3}{4\log 2 + \log 3} = 2bc$

$a^{x} = b \Rightarrow$

So, $x\log a = \log b$

$x = \frac{\log b}{\log a} = \log_{a}b$.