Solve 1/7 ln(x + 2)⁷ + 1/2 ln x - ln(x² + 3x + 2)²

Solution:

Given, 1/7ln((x + 2)⁷) + 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, 1/7ln((x + 2)⁷) + 1/2[ln x - ln((x²+ 3x + 2)²)] becomes

1/7ln((x + 2)⁷) + 1/2[ln x - ln((x + 1)(x + 2))²]

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

Now, \(\frac{1}{7}ln((x+2)^{7})\\=ln(((x+2)^{7})^{\frac{1}{7}})\\\\=ln(x+2)\)[Cancelling the common factor]

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

[Cancelling the common factor]

Using the property of logarithms, \(log_{b}(x)-log_{b}(y)=log_{b}(\frac{x}{y})\)

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

Now,\(\frac{1}{7}ln((x+2)^{7})+\frac{1}{2}(ln(x)-ln(((x+1)(x+2))^{2})\\=ln(x+2)+ln(\frac{x^{\frac{1}{2}}}{((x+1)(x+2))})\)

Using the property of logarithms, \(log_{b}x+log_{b}(y)=log_{b}(xy)\)

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

Therefore, \(\frac{1}{7}ln((x+2)^{7})+\frac{1}{2}[ln(x)-ln((x+1)(x+2))^{2}]=ln[\frac{x^{\frac{1}{2}}}{(x+1)}]\).