Express $\ln \sqrt[3]{32}$ in terms of $\ln2$ and/or $\ln3$
My try:
$\ln\sqrt[3]{32}= \ln(32^{1/3})= \frac{1}{3} \ln32 = \frac{1}{3} \ln2^5$
This isn't correct because there's a $5$...how do you solve this? Thank you.
$\endgroup$ 03 Answers
$\begingroup$Hint: How did you go from $\ln(32^{1/3})$ to $\frac{1}{3}\ln(32)$? Can you make the same step at the end?
$\endgroup$ 2 $\begingroup$$\ln(32^{1/3}) = 1/3(\ln 2^5 ) = \frac{5}{3} \ln 2$
$\endgroup$ $\begingroup$$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$
$$ \color{#0000ff}{\large\ln\pars{\root[{3 \atop}]{\vphantom{\large A}32}}} = \ln\pars{2^{5/3}} = {5 \over 3}\,\ln\left(2\right) =\color{#0000ff}{\large% \braces{{\ln\pars{2} \over \ln\pars{2}} + {\expo{\ln\pars{2}} \over \expo{\ln\pars{3}}}}\,\ln\left(2\right)} $$
$\endgroup$
