Question

# Write the expression as a sum and/or difference of logarithms. Express powers as factors. \ln \frac{2x\sqrt{1+8x}}{(x-3)^{13}}, x>3 \ln \frac{2x\sqrt{1+8x}}{(x-3)^{13}}=

Factors and multiples
Write the expression as a sum and/or difference of logarithms. Express powers as factors.
$$\displaystyle{\ln{{\frac{{{2}{x}\sqrt{{{1}+{8}{x}}}}}{{{\left({x}-{3}\right)}^{{{13}}}}}}}},{x}{>}{3}$$
$$\displaystyle{\ln{{\frac{{{2}{x}\sqrt{{{1}+{8}{x}}}}}{{{\left({x}-{3}\right)}^{{{13}}}}}}}}=$$ (Simplify your answer.)

2021-08-04
Step 1
The given expression is
$$\displaystyle{\ln{{\frac{{{2}{x}\sqrt{{{1}+{8}{x}}}}}{{{\left({x}-{3}\right)}^{{{13}}}}}}}},{x}{>}{3}$$
Step 2
Using the formula
$$\displaystyle{\ln{{\left({\frac{{{a}}}{{{b}}}}\right)}}}={\ln{{a}}}-{\ln{{b}}}$$
$$\displaystyle{\ln{{\left({a}{b}\right)}}}={\ln{{a}}}+{\ln{{b}}}$$
$$\displaystyle{{\ln{{a}}}^{{{n}}}=}{n}{\ln{{a}}}$$
Step 3
On simplifying with the help of given formula
$$\displaystyle{\ln{{\frac{{{2}{x}\sqrt{{{1}+{8}{x}}}}}{{{\left({x}-{3}\right)}^{{{13}}}}}}}}={\ln{{\frac{{{2}{x}{\left({1}+{8}{x}\right)}^{{{\frac{{{1}}}{{{2}}}}}}}}{{{\left({x}-{3}\right)}^{{{13}}}}}}}}$$
$$\displaystyle={\ln{{2}}}{x}{\left({1}+{8}{x}\right)}^{{{\frac{{{1}}}{{{2}}}}}}-{{\ln{{\left({x}-{3}\right)}}}^{{{13}}}}$$
$$\displaystyle={\ln{{2}}}{x}+{{\ln{{\left({1}+{8}{x}\right)}}}^{{{\frac{{{1}}}{{{2}}}}}}-}{{\ln{{\left({x}-{3}\right)}}}^{{{13}}}}$$
$$\displaystyle={\ln{{2}}}{x}+{{\ln{{\left({1}+{8}{x}\right)}}}^{{{\frac{{{1}}}{{{2}}}}}}-}{{\ln{{\left({x}-{3}\right)}}}^{{{13}}}}$$
$$\displaystyle={\ln{{2}}}+{\ln{{x}}}+{\frac{{{1}}}{{{2}}}}{\ln{{\left({1}+{8}{x}\right)}}}-{13}{\ln{{\left({x}-{3}\right)}}}$$
$$\displaystyle\Rightarrow{\ln{{\frac{{{2}{x}\sqrt{{{1}+{8}{x}}}}}{{{\left({x}-{3}\right)}^{{{13}}}}}}}}={\ln{{2}}}+{\ln{{x}}}+{\frac{{{1}}}{{{2}}}}{\ln{{\left({1}+{8}{x}\right)}}}-{13}{\ln{{\left({x}-{3}\right)}}}$$