Solution of Logarithmic Equations

$\log_6 (x - 2) + \log_6 (x + 3) = 1$

$\log_6 (x - 2)(x + 3) = \log_6 6$
 

Hence,
$(x - 2)(x + 3) = 6$

$x^2 + x - 6 = 6$

$x^2 + x - 12 = 0$

$(x + 4)(x - 3) = 0$

$x = -4 ~ \text{ and/or } ~ 3$
 

Check the given equation for x = -4

$\log_6 (x - 2) + \log_6 (x + 3) = 1$

$\log_6 (-4 - 2) + \log_6 (-4 + 3) = 1$

$\log_6 (-6) + \log_6 (-1) \ne 1$   ←   not okay

Check the given equation for x = 3

$\log_6 (x - 2) + \log_6 (x + 3) = 1$

$\log_6 (3 - 2) + \log_6 (3 + 3) = 1$

$\log_6 1 + \log_6 6 = 1$

$0 + 1 = 1$   ←   okay!

Thus, x = 3           answer

$x^{\log x} = 10\,000$

$\log x^{\log x} = \log 10\,000$

$\log x \cdot \log x = 4$

$(\log x)^2 = 4$

$\log x = \pm 2$

$x = 10^{\pm 2}$
 

Check the given equation for x = 10² = 100

$x^{\log x} = 10\,000$

$100^{\log 100} = 10\,000$

$100^2 = 10\,000$   ←   okay!

Check the given equation for x = 10^-2 = 1/100

$x^{\log x} = 10\,000$

$\left( \frac{1}{100} \right)^{\log (\frac{1}{100})} = 10\,000$

$\left( \frac{1}{100} \right)^{-2} = 10\,000$   ←   okay!

Hence, x = 100 and 1/100           answer