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First let's split this logarithm into two. We can use the property of logarithms, , to do this:
The second logarithm simplifies:
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:
If we can separate the 2's into separate logarithms, each of them would be an "x". We can use another property of logarithms,
The first three logs are x:
which simplifies to:
All we have left to do is to express 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. and . We can use this connection to take the equation for y and modify it to give us an expression for :
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, , which comes to our rescue. This lets us move a coefficient in front of the logarithm into the argument as an exponent! 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.
To introduce the coefficient of 1/2 all we need to do is multiply both sides of our equation by 1/2:
Now we can use the property:
and simplify:
Now we can return to our expression:
and substitute for :
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