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
ANSWERED
asked 2021-08-03
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.)

Answers (1)

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
\(\displaystyle{\ln{{\frac{{{2}{x}\sqrt{{{1}+{8}{x}}}}}{{{\left({x}-{3}\right)}^{{{13}}}}}}}}={\ln{{\frac{{{2}{x}{\left({1}+{8}{x}\right)}^{{{\frac{{{1}}}{{{2}}}}}}}}{{{\left({x}-{3}\right)}^{{{13}}}}}}}}\)
\(\displaystyle={\ln{{2}}}{x}{\left({1}+{8}{x}\right)}^{{{\frac{{{1}}}{{{2}}}}}}-{{\ln{{\left({x}-{3}\right)}}}^{{{13}}}}\)
\(\displaystyle={\ln{{2}}}{x}+{{\ln{{\left({1}+{8}{x}\right)}}}^{{{\frac{{{1}}}{{{2}}}}}}-}{{\ln{{\left({x}-{3}\right)}}}^{{{13}}}}\)
\(\displaystyle={\ln{{2}}}{x}+{{\ln{{\left({1}+{8}{x}\right)}}}^{{{\frac{{{1}}}{{{2}}}}}}-}{{\ln{{\left({x}-{3}\right)}}}^{{{13}}}}\)
\(\displaystyle={\ln{{2}}}+{\ln{{x}}}+{\frac{{{1}}}{{{2}}}}{\ln{{\left({1}+{8}{x}\right)}}}-{13}{\ln{{\left({x}-{3}\right)}}}\)
\(\displaystyle\Rightarrow{\ln{{\frac{{{2}{x}\sqrt{{{1}+{8}{x}}}}}{{{\left({x}-{3}\right)}^{{{13}}}}}}}}={\ln{{2}}}+{\ln{{x}}}+{\frac{{{1}}}{{{2}}}}{\ln{{\left({1}+{8}{x}\right)}}}-{13}{\ln{{\left({x}-{3}\right)}}}\)
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