Question 824341
{{{log(7, (x^2)) = p}}} and {{{log(7, (xy^2)) = q}}}
What we are going to do is...<ol><li>Solve the above equations to get expressions for {{{log(7, (x))}}} and {{{log(7, (y))}}}.</li><li>Rewrite {{{log(7, (root(3, xy)))}}} in terms of {{{log(7, (x))}}} and {{{log(7, (y))}}}.</li><li>Substitute in the expressions for {{{log(7, (x))}}} and {{{log(7, (y))}}} from step 1 into the expression we got in step 2.</li><li>Simplify.</li></ol>Solving {{{log(7, (x^2)) = p}}} for {{{log(7, (x))}}}:
Using a property of logarithms, {{{log(a, (j^n)) = n*log(a, (j))}}}, to move the exponent out in front:
{{{2log(7, (x)) = p}}}
Dividing both sides by 2:
{{{log(7, (x)) = p/2}}}<br>
Solving {{{log(7, (xy^2)) = q}}} for {{{log(7, (y))}}}:
Using another property of logarithms, {{{log(a, (j*k)) = log(a, (j)) + log(a, (k))}}}, we can split the argument into separate logs:
{{{log(7, (x)) + log(7, (y^2)) = q}}}
Substituting in the expression we got for {{{log(7, (x))}}}:
{{{p/2 + log(7, (y^2)) = q}}}
Using the property to move the exponent:
{{{p/2 + 2log(7, (y)) = q}}}
Multiplying each side by 1/2:
{{{p/4 + log(7, (y)) = q/2}}}
Subtracting p/4:
{{{log(7, (y)) = q/2-p/4}}}<br>
Rewriting {{{log(7, (root(3, xy)))}}} in terms of {{{log(7, (x))}}} and {{{log(7, (y))}}}:
Since a cube root is the same as an exponent of 1/3:
{{{log(7, ((xy)^(1/3)))}}}
Moving the exponent:
{{{(1/3)log(7, (xy))}}}
Split the argument:
{{{(1/3)(log(7, (x)) + log(7, (y)))}}}
Substituting in our expressions for the two logs:
{{{(1/3)(p/2 + q/2 - p/4)}}}
Simplifying...
{{{(1/3)(2p/4 + q/2 - p/4)}}}
{{{(1/3)(p/4 + q/2)}}}
{{{p/12 + q/6)}}}