Verified Answer :

apologiabiology

\(remember\)
\(ln(x)=log_e(x)\)
\(and\)
\(log_x(a)-log_x(b)= log_x( \frac{a}{b} )\)
\(and\)
\(log_a(b)=c\) means \(a^c=b\)
\(and\)
\(blog(a)=log(a^b)\)



\(so\)

\(ln(y+1)-ln(y)=1+3ln(x)\)
\(ln( \frac{y+1}{y} )=1+ln(x^3)\)

\(minus\) \(ln(x^3)\) \(from\) \(both\) \(sides\)

\(ln( \frac{y+1}{y}-ln(x^3) )=1\)
\(ln( \frac{y+1}{yx^3})=1\)
\(log_e( \frac{y+1}{yx^3})=1\)

\(translate\)

\(e^1=\frac{y+1}{yx^3}\)
\(e=\frac{y+1}{yx^3}\)

\(times\) \(both\) \(sides\) \(by\) \(yx^3\)

\(eyx^3=y+1\)

\(minus\) \(y\) \(from\) \(both\) \(sides\)

\(eyx^3-y=1\)
\(y(ex^3-1)=1\)

\(divide\) \(both\) \(sides\) \(by\) \((ex³-1)\)

\(y= \frac{1}{ex^3-1}\)

Applying logarithm properties, it is found that the correct expression is:

\(y = \frac{1}{ex^3 - 1}\)

These following properties are applied:

\(\ln{a} + \ln{b} = \ln{(ab)}\)

\(\ln{a} - \ln{b} = \ln{\left(\frac{a}{b}\right)}\)

\(a\ln{x} = \ln{x^a}\)

Also, since \(e^1 = e, 1 = \ln{e}\)

The original expression is:

\(\ln{(y + 1)} - \ln{y} = 1 + 3\ln{x}\)

Since \(1 = \ln{e}\)

\(\ln{(y + 1)} - \ln{y} = \ln{e} + 3\ln{x}\)

Applying the properties, we have that:

\(\ln{\left(\frac{y+1}{y}\right)} = \ln{e} + \ln{x^3}\)

\(\ln{\left(\frac{y+1}{y}\right)} = \ln{ex^3}\)

Removing the logarithms

\(\frac{y+1}{y} = ex^3\)

\(y + 1 = yex^3\)

\(yex^3 - y = 1\)

\(y(ex^3 - 1) = 1\)

\(y = \frac{1}{ex^3 - 1}\)

A similar problem is given at https://brainly.com/question/14405084

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