Answer:
Left-hand side: [tex]\displaystyle -\frac{1}{2}\, \ln(5)[/tex].
Right-hand side: [tex]\displaystyle \frac{1}{2}\, \ln(5) - \ln(5)[/tex].
Step-by-step explanation:
Apply the logarithm power rule: [tex]\ln \left(x^{a}\right) = a\, \ln (x)[/tex] for all [tex]x >0[/tex].
This property is not only true for logarithm to the base [tex]e[/tex], but for other bases, as well.
Take the logarithm (to the base [tex]e[/tex]) of the left-hand side of this equation:
[tex]\displaystyle \ln \left(5^{-1/2}\right) = (-1/2)\, \ln(5)[/tex].
For the right-hand side of this equation, consider the logarithm quotient rule:
[tex]\displaystyle \ln \left(\frac{a}{b}\right) = \ln(a) - \ln (b)[/tex] for all [tex]a> 0[/tex] and [tex]b > 0[/tex].
Indeed, on the right-hand side of this equation, [tex]\sqrt{5} > 0[/tex] and [tex]5 > 0[/tex]. Therefore:
[tex]\displaystyle \ln\left(\frac{\sqrt{5}}{5}\right) = \ln\left(\sqrt{5}\right) - \ln(5)[/tex].
This expression could be further simplified. Notice that [tex]\sqrt{x}[/tex] is equivalent to [tex]x^{1/2}[/tex] for all [tex]x \ge 0[/tex]. (Think about how [tex]\sqrt{x} \cdot \sqrt{x} =x[/tex] whereas [tex]x^{1/2} \cdot x^{1/2} = x^{(1/2) + (1/2)} = x[/tex].)
Therefore, [tex]\ln \left(\sqrt{5}\right)[/tex] would be equivalent to [tex]\ln\left(5^{1/2}\right)[/tex]. Apply the logarithm power rule to show that [tex]\displaystyle \ln\left(5^{1/2}\right) = \frac{1}{2}\, \ln(5)[/tex].
[tex]\begin{aligned} \text{R.H.S.} &= \ln\left(\frac{\sqrt{5}}{5}\right) \\ &= \ln\left(\sqrt{5}\right) - \ln(5) \\ &= \frac{1}{2}\, \ln(5) - \ln (5) = -\frac{1}{2}\, \ln(5)\end{aligned}[/tex].
Indeed, the left-hand side of this equation matches the right-hand side.
