# Use the properties of logarithms to rewrite each expression as the logarithm of

Use the properties of logarithms to rewrite each expression as the logarithm of a single expression. Be sure to use positive exponents and avoid radicals.

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Aniqa O'Neill
Using the above properties of logarithm, we get
$$\displaystyle{2}{\ln{{4}}}{x}^{{3}}+{3}{\ln{{y}}}-{\frac{{{1}}}{{{3}}}}{{\ln{{z}}}^{{6}}}$$
$$\displaystyle={{\ln{{\left({4}{x}^{{3}}\right)}}}^{{2}}+}{{\ln{{\left({y}\right)}}}^{{3}}-}{{\ln{{\left({z}^{{6}}\right)}}}^{{\frac{{{1}}}{{{3}}}}}}$$
$$\displaystyle={\ln{{16}}}{x}^{{6}}+{{\ln{{y}}}^{{3}}-}{{\ln{{z}}}^{{2}}}$$
$$\displaystyle={\ln{{16}}}{x}^{{6}}+{\ln{{\frac{{{y}^{{3}}}}{{{z}^{{2}}}}}}}$$
$$\displaystyle={\ln{{\left({16}{x}^{{6}}\times{\frac{{{y}^{{3}}}}{{{z}^{{2}}}}}\right)}}}$$
$$\displaystyle={\ln{{\frac{{{16}{x}^{{6}}{y}^{{3}}}}{{{z}^{{2}}}}}}}$$
$$\displaystyle\therefore{2}{\ln{{4}}}{x}^{{3}}+{3}{\ln{{y}}}-{\frac{{{1}}}{{{3}}}}{{\ln{{z}}}^{{6}}=}{\ln{{\frac{{{16}{x}^{{6}}{y}^{{3}}}}{{{z}^{{2}}}}}}}$$