Question

Write in logarithmic form: B5%7D" id="TexFormula1" title="5^\frac{-1}{2} = \frac{\sqrt{5} }{5}" alt="5^\frac{-1}{2} = \frac{\sqrt{5} }{5}" align="absmiddle" class="latex-formula">

Answer 1

Answer:

Left-hand side: $\displaystyle -\frac{1}{2}\, \ln(5)$.

Right-hand side: $\displaystyle \frac{1}{2}\, \ln(5) - \ln(5)$.

Step-by-step explanation:

Apply the logarithm power rule: $\ln \left(x^{a}\right) = a\, \ln (x)$ for all $x >0$.

This property is not only true for logarithm to the base $e$, but for other bases, as well.

Take the logarithm (to the base $e$) of the left-hand side of this equation:

$\displaystyle \ln \left(5^{-1/2}\right) = (-1/2)\, \ln(5)$.

For the right-hand side of this equation, consider the logarithm quotient rule:

$\displaystyle \ln \left(\frac{a}{b}\right) = \ln(a) - \ln (b)$ for all $a> 0$ and $b > 0$.

Indeed, on the right-hand side of this equation, $\sqrt{5} > 0$ and $5 > 0$. Therefore:

$\displaystyle \ln\left(\frac{\sqrt{5}}{5}\right) = \ln\left(\sqrt{5}\right) - \ln(5)$.

This expression could be further simplified. Notice that $\sqrt{x}$ is equivalent to $x^{1/2}$ for all $x \ge 0$. (Think about how $\sqrt{x} \cdot \sqrt{x} =x$ whereas $x^{1/2} \cdot x^{1/2} = x^{(1/2) + (1/2)} = x$.)

Therefore, $\ln \left(\sqrt{5}\right)$ would be equivalent to $\ln\left(5^{1/2}\right)$. Apply the logarithm power rule to show that $\displaystyle \ln\left(5^{1/2}\right) = \frac{1}{2}\, \ln(5)$.

$\begin{aligned} \text{R.H.S.} &= \ln\left(\frac{\sqrt{5}}{5}\right) \\ &= \ln\left(\sqrt{5}\right) - \ln(5) \\ &= \frac{1}{2}\, \ln(5) - \ln (5) = -\frac{1}{2}\, \ln(5)\end{aligned}$.

Indeed, the left-hand side of this equation matches the right-hand side.