Verified Answer :

First, it is important that you know the rules of logs

So that is blog(a)=log(ab)

So you will get ln(z^(1/2))-ln(5-x)-ln(y^4)

You can also
do it this way, applying another rule of logs

og(a)−log(b)=log(ab)

Then substitute it to the given, you get:

ln(z1/2)−ln(5+x)−ln(y4)

ln[z1/2(5+x)y4]

Answer:

The answer is:

\(ln(\frac{z^\frac{1}{2}}{(5+x)*(y^4)} )\)

Step-by-step explanation:

In order to determine the answer, we have to know about the properties of the logarithmic.

For this case, we need three properties. It is very important that these properties assume that the base of the logarithmic is the same. In this case we have the base "e" (euler number).

  1. We can separate the power inside of the logarithmic:                  \(ln(a^b)=b*ln(a)\).
  2. We can join the sum of logarithmics into a product:                          \(ln(a)+ln(b)=ln(a*b)\)
  3. We can join the subtraction of logarithmics into a division:                     \(ln(a)-ln(b)=ln(\frac{a}{b})\)

Applying these properties:

\(\frac{1}{2}ln(z)=ln(z^\frac{1}{2})\\\\4ln(y)=ln(y^4)\\\\ln(z^\frac{1}{2})-ln(5+x)-ln(y^4)\\ln(z^\frac{1}{2})-(ln(5+x)+ln(y^4))\\ln(z^\frac{1}{2})-(ln((5+x)*(y^4)))\\ln(\frac{z^\frac{1}{2}}{(5+x)*(y^4)} )\)

Finally, the expression is:

\(ln(\frac{z^\frac{1}{2}}{(5+x)*(y^4)} )\)

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