Dachekar Mombrun

2022-07-05

Nick Camelot

The given expression is:
$\mathrm{ln}\left(\frac{\left(8x{\right)}^{\frac{1}{11}}\left(1+4x\right)}{\left(x-6{\right)}^{9}}\right)$
First, let's simplify the expression inside the natural logarithm:
$\left(8x{\right)}^{\frac{1}{11}}=\left(\left({2}^{3}x{\right)}^{\frac{1}{11}}\right)={2}^{\frac{3}{11}}{x}^{\frac{1}{11}}$
$\left(x-6{\right)}^{9}=\left(-\left(6-x\right){\right)}^{9}=\left(-1{\right)}^{9}\left(6-x{\right)}^{9}=-\left(6-x{\right)}^{9}$
Now, we can rewrite the original expression using the properties of logarithms:
$\mathrm{ln}\left(\frac{\left(8x{\right)}^{\frac{1}{11}}\left(1+4x\right)}{\left(x-6{\right)}^{9}}\right)=\mathrm{ln}\left({2}^{\frac{3}{11}}{x}^{\frac{1}{11}}\left(1+4x\right)\right)-\mathrm{ln}\left(\left(6-x{\right)}^{9}\right)$
Next, we can further simplify each logarithm:
$\mathrm{ln}\left({2}^{\frac{3}{11}}{x}^{\frac{1}{11}}\left(1+4x\right)\right)=\frac{3}{11}\mathrm{ln}\left(2\right)+\frac{1}{11}\mathrm{ln}\left(x\right)+\mathrm{ln}\left(1+4x\right)$
$\mathrm{ln}\left(\left(6-x{\right)}^{9}\right)=9\mathrm{ln}\left(6-x\right)$
Finally, we have the expression as the sum and/or difference of logarithms:
$\mathrm{ln}\left(\frac{\left(8x{\right)}^{\frac{1}{11}}\left(1+4x\right)}{\left(x-6{\right)}^{9}}\right)=\frac{3}{11}\mathrm{ln}\left(2\right)+\frac{1}{11}\mathrm{ln}\left(x\right)+\mathrm{ln}\left(1+4x\right)-9\mathrm{ln}\left(6-x\right)$
Therefore, the expression can be written as the sum and/or difference of logarithms:
$\mathrm{ln}\left(\frac{\left(8x{\right)}^{\frac{1}{11}}\left(1+4x\right)}{\left(x-6{\right)}^{9}}\right)=\frac{3}{11}\mathrm{ln}\left(2\right)+\frac{1}{11}\mathrm{ln}\left(x\right)+\mathrm{ln}\left(1+4x\right)-9\mathrm{ln}\left(6-x\right)$

Do you have a similar question?