SOLUTION: .｝log(x-5y)=2 logx-logy=1

Question 238185: .｝log(x-5y)=2
logx-logy=1

Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!

I assume the problem is to solve this system of equations. Since the variables are in the arguments of logarithms we will rewrite these equations in exponential form. However, the second equation is not in a proper form for this yet so we have to start there.

We need to "combine" the logarithms of the second equation using this property of logarithms: :
Now, with our modified second equation our system is:

These are now ready to be rewritten in exponential form using the fact that is equivalent to :

Simplifying the powers of 10 we get:

The second equation is not linear. So the Substitution Method is best. By adding 5y to the first equation we can solve it for x:

And now we can substitute this into the second equation giving:

To solve this we eliminate the fraction by multiplying both sides by y:

Subtract 5y from each side:

Divide both sides by 5:

Then we can use this to find x, using x = 5y+100:

So we have a solution: (200, 20)

With logarithm problems it is important to check answers because we cannot allow solutions which make arguments of logarithms zero or negative. So we will substitute these values into the original equations to see if they actually work:

We can see that the arguments of all three logarithms are positive. So our solution works.
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SOLUTION: .&#65373;log(x-5y)=2 logx-logy=1

SOLUTION: .｝log(x-5y)=2 logx-logy=1