Question 275180
{{{log(a, (40/(a^3)))}}}
First let's split this logarithm into two. We can use the property of logarithms, {{{log(a, (p/q)) = log(a, (p)) - log(a, (q))}}}, to do this:
{{{log(a, (40)) - log(a, (a^3))}}}
The second logarithm simplifies:
{{{log(a, (40)) - 3}}}
Now we need to express the remaining logarithm in terms of x and y. If we factor 40 we might be able to see a path to our solution:
{{{log(a, (2*2*2*5)) - 3}}}
If we can separate the 2's into separate logarithms, each of them would be an "x". We can use another property of logarithms, {{{log(a, (p*q)) = log(a, (p)) + log(a, (q))}}, to do exactly this:
{{{log(a, (2)) + log(a, (2)) + log(a, (2)) + log(a, (5)) - 3}}}
The first three logs are x:
{{{x + x + x + log(a, (5)) - 3}}}
which simplifies to:
{{{3x + log(a, (5)) - 3}}}
All we have left to do is to express {{{log(a, (5))}}} in terms of x and/or y. Since 2 is not a factor of 5 it would seem that x will not help us with this. 25 is not a factor of 5 either so at first glance it would seem that y will not help either. But there is a connection between 5 and 25. {{{5^2 = 25}}} and {{{sqrt(25) = 5}}}. We can use this connection to take the equation for y and modify it to give us an expression for {{{log(a, (5))}}}:
{{{y = log(a, (25))}}}
Square roots are exponents of 1/2. So somehow we need to get an exponent of 1/2 on the 25. The only legitimate way is a little tricky. We cannot just raise both sides to the 1/2 power. This would put the exponent on the logarithm, not on the 25 in the argument of the logarithm. But we have yet another property of logarithms, {{{q*log(a, (p)) = log(a, (p^q))}}}, which comes to our rescue. This lets us move a coefficient in front of the logarithm <b>into the argument as an exponent!</b> So if we can introduce a coefficient of 1/2 we can then use this property to move it into the argument as the exponent of 25.<br>
To introduce the coefficient of 1/2 all we need to do is multiply both sides of our equation by 1/2:
{{{(1/2)y = (1/2)log(a, (25))}}}
Now we can use the property:
{{{(1/2)y = log(a, (25^(1/2)))}}}
and simplify:
{{{(1/2)y = log(a, (5))}}}
Now we can return to our expression:
{{{3x + log(a, (5)) - 3}}}
and substitute for {{{log(a, (5))}}}:
{{{3x + (1/2)y - 3}}}