if log xy=6 and log x/y=2. solve for x and y

Answer my question

Y =4

To solve for x and y in the given equations, we can use properties of logarithms. Let's start by solving for x in the equation log(xy) = 6.

Using the property log(xy) = log x + log y, we can rewrite the equation as:
log x + log y = 6

Next, let's solve for y in terms of x:
log y = 6 - log x

To eliminate the logarithm, we can take the exponential of both sides using the base 10. Remember that taking the exponential of a logarithmic equation cancels out the logarithm.

10^(log y) = 10^(6 - log x)

This simplifies to:
y = 10^(6 - log x)

Now, let's move on to the second equation log(x/y) = 2.

Using the property log(x/y) = log x - log y, we can rewrite the equation as:
log x - log y = 2

Next, let's solve for y in terms of x:
log y = log x - 2

Taking the exponential of both sides using the base 10, we get:
y = 10^(log x - 2)

Now we have expressions for y in terms of x in both equations. We can substitute these expressions into each other to solve for x.

Substituting y = 10^(6 - log x) into the equation y = 10^(log x - 2):
10^(6 - log x) = 10^(log x - 2)

Since the bases are the same, the exponents must be equal:
6 - log x = log x - 2

Adding log x to both sides:
6 = 2log x - 2

Adding 2 to both sides:
8 = 2log x

Dividing by 2:
4 = log x

To solve for x, we need to evaluate 10^(log x) = 10^4:
x = 10^4
x = 10000

Now, let's substitute x = 10000 into the expression for y:

y = 10^(6 - log x)
y = 10^(6 - log 10000)

Since log 10000 = 4 (10^4 = 10000), we have:
y = 10^(6 - 4)
y = 10^2
y = 100

Therefore, the solution to the equations log(xy) = 6 and log(x/y) = 2 is x = 10000 and y = 100.

logx + logy = 6

logx - logy = 2

Now just solve as usual for logx and logy.

Then raise whatever base you are using (probably 10) to the respective powers.