Solve 1/5ln((x + 2)⁵) + 1/2[ln x - ln((x² + 3x + 2)²)]

Solution:

Given, 1/5ln((x + 2)⁵) + 1/2[ln x - ln((x² + 3x + 2)²)]

We have to solve the equation.

First, find the factors of (x² + 3x + 2)

x² + 3x + 2 = x² + 2x + x + 2

=x(x + 2) + 1(x + 2)

= (x + 1)(x + 2)

So, 1/5ln((x + 2)⁵) + 1/2[ln x - ln((x² + 3x + 2)²)] becomes 1/5ln((x + 2)⁵) + 1/2[ln x - ln((x + 1)(x + 2))²]

We know, \(b\, log_{a}=loga^{b}\)

Now, \(\frac{1}{5}ln((x+2)^{5})=ln(((x+2)^{5})^{\frac{1}{5}})\)

Cancelling the common factor,

\(\frac{1}{5}ln((x+2)^{5})=ln(x+2)\)

Similarly, \(\frac{1}{2}[ln(x)-ln((x+1)(x+2))^{2}]=\frac{1}{2}ln(x)-\frac{1}{2}ln((x+1)(x+2))^{2}\)

\(\frac{1}{2}ln(x)-\frac{1}{2}ln((x+1)(x+2))^{2}=[ln(x^{\frac{1}{2}})]-[ln(((x+1)(x+2))^{2})^{\frac{1}{2}}]\)

Cancelling the common factor,

\([ln(x^{\frac{1}{2}})]-[ln(((x+1)(x+2))^{2})^{\frac{1}{2}}]=[ln(x)^{\frac{1}{2}}-ln((x+1)(x+2))]\)

Using the property of logarithms,

\(log_{b}(x)-log_{b}(y)=log_{b}(\frac{x}{y})\)

So,\([ln(x)^{\frac{1}{2}}-ln((x+1)(x+2))]=ln(\frac{x^{\frac{1}{2}}}{((x+1)(x+2))})\)

Now,\(\frac{1}{5}ln((x+2)^{5})+\frac{1}{2}(ln(x)-ln(((x+1)(x+2))^{2})=ln(x+2)+ln(\frac{x^{\frac{1}{2}}}{((x+1)(x+2))})\)

Using the property of logarithms,

\(log_{b}x+log_{b}(y)=log_{b}(xy)\)

So, \(ln(x+2)+ln(\frac{x^{\frac{1}{2}}}{((x+1)(x+2))})=ln[(x+2)(\frac{x^{\frac{1}{2}}}{((x+1)(x+2))})]\)

\(ln[(x+2)(\frac{x^{\frac{1}{2}}}{((x+1)(x+2))})]=ln[\frac{x^{\frac{1}{2}}}{(x+1)}]\)

Therefore, \(\frac{1}{5}ln((x+2)^{5})+\frac{1}{2}[ln(x)-ln((x+1)(x+2))^{2}]=ln[\frac{x^{\frac{1}{2}}}{(x+1)}]\).

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Solve 1/5ln((x + 2)⁵) + 1/2[ln x - ln((x² + 3x + 2)²)]

Summary:

\(\frac{1}{5}ln((x+2)^{5})+\frac{1}{2}[ln(x)-ln((x+1)(x+2))^{2}]=ln[\frac{x^{\frac{1}{2}}}{(x+1)}]\).