Solve simultaneously log (x-1)+2logy=2log3 and logx+logy=log6
log ( x -1 ) + 2 log y = 2 log 3
log ( x -1 ) = 2 log 3 - 2 log y
log ( x -1 ) = 2 ( log 3 - log y )
log ( x -1 ) = 2 log ( 3 / y )
log ( x -1 ) = log [ ( 3 / y )² ]
x - 1 = ( 3 / y )²
x - 1 = 9 / y²
x = 9 / y² + 1
log x + log y = log 6
log ( x ∙ y ) = log 6
x ∙ y = 6
x = 6 / y
Use fact:
x = x
9 / y² + 1 = 6 / y
Multiply both sides by y²
9 + y² = 6 y
Subtract 6 y to both sides
9 + y² - 6 y = 6 y - 6 y
y² - 6 y + 9 = 0
Factor the left hand side
( y - 3 )² = 0
y - 3 = √0
y - 3 = 0
add 3 to both sides
y - 3 + 3 = 0 + 3
y = 3
x = 6 / y
x = 6 / 3
x = 2
Solution x = 2 , y = 3
Proof:
log ( x -1 ) + 2 log y = 2 log 3
log ( 2 -1 ) + 2 log 3 = 2 log 3
log 1 + 2 log 3 = 2 log 3
0 + 2 log 3 = 2 log 3
2 log 3 = 2 log 3
log x + log y = log 6
log 2 + log 3 = log 6
log ( 2 ∙ 3 ) = log 6
log 6 = log 6
To solve the simultaneous equations
1. log(x - 1) + 2log(y) = 2log(3) (Equation 1)
2. log(x) + log(y) = log(6) (Equation 2)
Step 1: Use the properties of logarithms to rewrite the equations.
For Equation 1:
log(x - 1) + log(y^2) = log(3^2)
log(x - 1) + log(y^2) = log(9)
For Equation 2:
log(x * y) = log(6)
Step 2: Apply the logarithmic rule log(a) + log(b) = log(a * b) to Equation 1.
log((x - 1) * y^2) = log(9)
Step 3: Equate the expressions inside the logarithm in Equation 2 with Equation 1.
(x - 1) * y^2 = 9
Step 4: Rewrite Equation 2 using the logarithmic rule log(a * b) = log(a) + log(b).
log(x) + log(y) = log(6)
log(x * y) = log(6)
Step 5: Equate the expressions inside the logarithm in Equation 2 with Equation 4.
(x * y) = 6
Step 6: Solve the simultaneous equations by substituting Equation 5 into Equation 3.
(x - 1) * y^2 = 9
=> (6 - 1) * y^2 = 9 (substituting x = 6 from Equation 5)
=> 5 * y^2 = 9
=> y^2 = 9/5
=> y = ±√(9/5) (taking the square root of both sides)
Step 7: Substitute the value of y into Equation 5.
(x * y) = 6
=> x * ±√(9/5) = 6
=> x = 6 / ±√(9/5) (dividing both sides by ±√(9/5))
Simplifying the denominator:
x = 6 / ±(3/√5)
=> x = 6 * ±(√5/3) (multiplying by √5/3 to rationalize the denominator)
Therefore, the solutions to the simultaneous equations are:
x = 6 * ±(√5/3)
y = ±√(9/5)
To solve the given system of equations simultaneously, we'll use logarithmic properties and algebraic manipulation. Here's how you can solve it step by step:
Equation 1: log(x - 1) + 2log(y) = 2log(3)
Equation 2: log(x) + log(y) = log(6)
Step 1: Simplify Equation 1 using logarithmic properties.
- Applying the power rule of logarithms, we can rewrite the equation as:
log((x - 1) * y^2) = log(3^2)
- Since the bases on both sides are the same (logarithm base 10), we can eliminate the logarithm and equate the arguments:
(x - 1) * y^2 = 9 ---- (Equation 1')
Step 2: Simplify Equation 2 using logarithmic properties.
- We can combine the logarithms on the left side using the product rule:
log(x * y) = log(6)
- Again, equating the arguments:
x * y = 6 ---- (Equation 2')
Step 3: Now, we have a system of equations:
(x - 1) * y^2 = 9 ---- (Equation 1')
x * y = 6 ---- (Equation 2')
Step 4: Solve the system of equations. We can choose either substitution or elimination method. Let's use the substitution method:
From Equation 2', we can express x in terms of y:
x = 6 / y
Substitute this value of x in Equation 1':
((6 / y) - 1) * y^2 = 9
(6 - y) * y^2 = 9y
Rearrange the equation:
6y^2 - y^3 = 9y
- y^3 + 6y^2 - 9y = 0
Step 5: Factor the equation (if possible) to find the solutions.
While factoring is not always possible for cubic equations, we can see that -y^3 + 6y^2 - 9y = 0 has a common factor of y.
y(-y^2 + 6y - 9) = 0
Factoring the quadratic expression inside the parentheses:
y(-1)(y - 3)(y - 3) = 0
The equation can be simplified to:
y(y - 3)^2 = 0
Therefore, the solutions for y are:
y = 0 (y = 0 implies x = 6 / 0, which is undefined)
y - 3 = 0 (y = 3 implies x = 6 / 3 = 2)
Step 6: Find the corresponding x-values using Equation 2'.
For y = 0, x is undefined.
For y = 3, x = 2.
Thus, the solution to the system of equations is x = 2 and y = 3.