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Write in logarithmic form:
5^(-1)/(2) = (√(5) )/(5)

User Nicktmro
by
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2 votes

Answer:

Left-hand side:
\displaystyle -(1)/(2)\, \ln(5).

Right-hand side:
\displaystyle (1)/(2)\, \ln(5) - \ln(5).

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((a)/(b)\right) = \ln(a) - \ln (b) for all
a> 0 and
b > 0.

Indeed, on the right-hand side of this equation,
√(5) > 0 and
5 > 0. Therefore:


\displaystyle \ln\left((√(5))/(5)\right) = \ln\left(√(5)\right) - \ln(5).

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

Therefore,
\ln \left(√(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) = (1)/(2)\, \ln(5).


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

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

User Mike Sherov
by
8.8k points

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