First slide

Types of real valued functions XI

Question

The equation $\log_{x + 1} (x - 0 .5) = \log_{x - 0 .5} (x + 1)$ has

Moderate

A

two real solutions
B

no prime solution
C

one integral solution
D

no irrational solution

Solution

$\begin{array}{l} \frac{\log_{2} (x - 0 .5)}{\log_{2} (x + 1)} = \frac{\log_{2} (x + 1)}{\log_{2} (x - 0 .5)} \\ or {[\log_{2} (x + 1)]}^{2} = {[\log_{2} (x - 0 .5)]}^{2} \\ ∴ \log_{2} (x + 1) = \pm \log_{2} (x - 0 .5) \\ If \log_{2} (x + 1) = \log_{2} (x - 0 .5) . Then, \\ x + 1 = x - 0 .5, hence no solution \\ If \log_{2} (x + 1) = \log (x - 0 .5)^{- 1} . Then, \\ x + 1 = \frac{1}{x - (1 / 2)} = \frac{2}{2 x - 1} \\ or (x + 1) (2 x - 1) = 2 \\ or 2 x^{2} + x - 3 = 0 \\ or 2 x^{2} + 3 x - 2 x - 3 = 0 \\ or (x - 1) (2 x + 3) = 0 \\ \Rightarrow x = 1 \end{array}$

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