Question 603761
{{{expr(2/3)*log((x))}}} + {{{expr(1/3)log((xy^3))}}}
<pre>
On both terms term use the rule {{{A*log((B))}}} = {{{log((B^A))}}}
to write the coefficients of the logs as exponents

{{{log((matrix(2,1,"",x^(2/3))))}}} + {{{log((matrix(2,1,"",(xy^3)^(1/3))))}}}

Now we simplify {{{matrix(2,1,"",(xy^3)^(1/3))}}} this way: {{{matrix(2,1,"",(xy^3)^(1/3))) }}} = {{{matrix(2,1,"",(x^1y^3)^(1/3))}}} = {{{matrix(2,1,"",    x^(1/3)y^(3/3)       )}}} = {{{matrix(2,1,"",    x^(1/3)y^1       )}}} ={{{matrix(2,1,"",    x^(1/3)y       )}}}

And the above becomes:

{{{log((matrix(2,1,"",x^(2/3))))}}} + {{{log((matrix(2,1,"",    x^(1/3)y       )))}}}

Now use the rule: {{{log((A)) + log((B))}}} = {{{log((A*B))}}}
and you have this:

{{{log((matrix(2,1,"",    x^(2/3)x^(1/3)y       )))}}}

Now we multiply the powers of x by adding the exponents of x: {{{2/3+1/3=3/3=1}}}

{{{log((x^1y))}}}

And the final answer is:

{{{log((xy))}}}

Edwin</pre>