Solve the simultenous equation.
XY=160
LogX - 3LogY=1
xy = 160
logx - 3logy = -1
logx + logy = log160
logx - 3logy = -1
subtract to get
4logy = log160 + 1 = log160 + log10 = log1600 = 2log40
logy = 1/2 log40
y = √40
so, x=160/√40 = 4√40
or,
xy = 160
logx - 3logy = -1
xy = 160
x/y^3 = 1/10
x = 160/y, so
160/y^4 = 1/10
y^4 = 1600
y = √40
x = 160/√40 = 4√40
yeah, I know √40 = 2√10 but I think it's easier to see this way.
To solve the simultaneous equation:
1. Start by expressing the second equation in terms of a single logarithm. Since the given equation has a difference of logarithms, you can rewrite it using logarithmic identities.
LogX - 3LogY = 1
Applying the power rule of logarithms, we can rewrite 3LogY as Log(Y^3):
LogX - Log(Y^3) = 1
Using the quotient rule of logarithms, we can combine both logs into a single logarithm:
Log(X / (Y^3)) = 1
2. To eliminate the logarithm, we can rewrite the equation in exponential form. Remember that log(base b) of x equals y can be expressed as b^y = x.
Exponentiating both sides of the equation with base 10:
10^1 = X / (Y^3)
Therefore, we have X / (Y^3) = 10.
3. Now, use the first equation XY = 160 and substitute the value of X from step 2:
(10)(Y^3) = 160
Divide both sides of the equation by 10:
Y^3 = 16
4. Take the cube root of both sides of the equation to solve for Y:
Y = ∛16
Y = 2
5. Substitute the value of Y back into the equation XY = 160 to solve for X:
X(2) = 160
X = 160 / 2
X = 80
Therefore, the solution to the simultaneous equation is X = 80 and Y = 2.