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KATRIN_1 [288]
3 years ago
8

Expanding Logarithmic Expressions In Exercise, use the properties of logarithms to rewrite the expression as a sum, difference,

or multiple of logarithms.
In (x^2(x - 1))1/2
Mathematics
1 answer:
aev [14]3 years ago
6 0

Answer:

The simplified expression is:

\ln{x} + \frac{\ln{(x-1)}}{2}

Step-by-step explanation:

We have those following logarithmic properties:

\ln{\frac{a}{b}} = \ln{a} - \ln{b}

\ln{a*b} = \ln{a} + \ln{b}

\ln{a^{n}} = n\ln{a}

In this problem, we have that:

\ln{x^{2}(x-1)}^{1/2}

Applying these properties

\frac{1}{2}*(\ln{x^{2} + \ln{(x-1)})

\ln{x} + \frac{\ln{(x-1)}}{2}

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A 41 gram sample of a substance that's used to detect explosives has a k-value of 0.1392. Find the substance's half life, in day
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Answer:

The substance half-life is of 4.98 days.

Step-by-step explanation:

Equation for an amount of a decaying substance:

The equation for the amount of a substance that decay exponentially has the following format:

A(t) = A(0)e^{-kt}

In which k is the decay rate, as a decimal.

k-value of 0.1392.

This means that:

A(t) = A(0)e^{-0.1392t}

Find the substance's half life, in days.

This is t for which A(t) = 0.5A(0). So

A(t) = A(0)e^{-0.1392t}

0.5A(0) = A(0)e^{-0.1392t}

e^{-0.1392t} = 0.5

\ln{e^{-0.1392t}} = \ln{0.5}

-0.1392t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.1392}

t = 4.98

The substance half-life is of 4.98 days.

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