Question

Solve $\log (a+\sqrt{a^2+1})+\log \left(\frac{1}{a+\sqrt{a^2+1}}\right)$ =

• A. $1$
• B. $2\log (a+\sqrt{a^2+1})$
• C. $0$
• D. none

Answer 1

Answer: (C)

$0$

Apply the property:

$\log a+\log b=\log ab$

Now,

$\log (a+\sqrt{a^2+1})+\log \left(\frac{1}{a+\sqrt{a^2+1}}\right)$

$=\log \left(\right(a+\sqrt{a^2+1})\times \frac{1}{a+\sqrt{a^2+1}})$

$=\log 1$

$=0$