Question 815429
{{{e^(9-6ln (x)+ln (y))}}}
Since {{{e^ln(q) = q}}} we will be able to solve this problem if we change the 9 into a logarithm and then combine the logs. Since ln(e) = 1 we can rewrite the expression:
{{{e^(9ln(e)-6ln (x)+ln (y))}}}
Now we can use the {{{n*log(a, (p)) = log(a, (p^n))}}} property to move the two coefficients into the argument as its exponent:
{{{e^(ln(e^9)-ln (x^6)+ln (y))}}}
Now we start combining the logs. For the first two we will use the {{{log(a, (p))-log(a, (q)) = log(a, (p/q))}}} property:
{{{(e^(ln(e^9/x^6)+ln (y)))}}}
For the remaining logs we will use the {{{log(a, (p))+log(a, (q)) = log(a, (p*q))}}} property:
{{{e^(ln((e^9/x^6)*y))}}}
We now have the desired "e to the ln" form. So this equals:
{{{(e^9/x^6)*y)}}}
or
{{{y*e^9/x^6}}}