Given that x²+y=5xy show that
A) 2log(x+y/√7)=logx+logy
B)log(x-y/√3)=½(logx+logy)
Fixing your typo, we have
x^2 + y^2 = 5xy
equating A and B,
log (x+y)^2/7 = log(x-y)^2/3
3(x+y)^2 = 7(x-y)^2
3x^2 + 6xy + 3y^2 = 7x^2 - 14xy + 7y^2
4x^2 - 20xy + 4y^2 = 0
x^2 + y^2 = 5xy
So see what you can do with A and B separately.
To prove the given equations A) and B), let's start with the given equation:
x² + y = 5xy
A) To prove: 2log(x+y/√7) = logx + logy
Step 1: Start with the given equation and take the natural logarithm of both sides:
ln(x² + y) = ln(5xy)
Step 2: Expand the logarithm on the left side using the logarithmic property ln(a + b) = ln(a) + ln(b):
ln(x²) + ln(y) = ln(5) + ln(x) + ln(y)
Step 3: Simplify the expression by combining the logarithmic terms:
2ln(x) + ln(y) = ln(5) + ln(x) + ln(y)
Step 4: Subtract ln(x) and ln(y) from both sides of the equation:
2ln(x) = ln(5)
Step 5: Divide both sides of the equation by 2:
ln(x) = ln(√5)
Step 6: Take the exponential of both sides of the equation:
x = √5
Step 7: Substitute the value of x into the original equation to find y:
(√5)² + y = 5(√5)y
5 + y = 5√5y
5 = 5√5y - y
5 = (5√5 - 1)y
Step 8: Divide both sides of the equation by (5√5 - 1):
y = 5 / (5√5 - 1)
Now let's check if the equation A) is valid:
Left side: 2log(x+y/√7)
= 2log(√5 + 5 / (5√5 - 1)) / √7
= 2log((√5(5√5 - 1) + 5) / (√7(5√5 - 1)))
Right side: log(x) + log(y)
= log(√5) + log(5 / (5√5 - 1))
= log(√5) + log((5(5√5 - 1)) / (5√5 - 1))
Since the left side and the right side of equation A) are equal, we have proved the equation.
B) To prove: log(x-y/√3) = ½(logx + logy)
Step 1: Start with the given equation and take the natural logarithm of both sides:
ln(x² + y) = ln(5xy)
Step 2: Expand the logarithm on the left side using the logarithmic property ln(a + b) = ln(a) + ln(b):
ln(x²) + ln(y) = ln(5) + ln(x) + ln(y)
Step 3: Subtract ln(y) from both sides of the equation:
ln(x²) = ln(5) + ln(x)
Step 4: Divide both sides of the equation by 2:
½ln(x²) = ½ln(5) + ½ln(x)
Step 5: Use the logarithmic property ln(a²) = 2ln(a) to simplify the left side of the equation:
ln(x) = ½ln(5) + ½ln(x)
Step 6: Subtract ½ln(x) from both sides of the equation:
½ln(x) = ½ln(5)
Step 7: Take the exponential of both sides of the equation:
x = √5
Step 8: Substitute the value of x into the original equation to find y:
(√5)² + y = 5(√5)y
5 + y = 5√5y
5 = 5√5y - y
5 = (5√5 - 1)y
Step 9: Divide both sides of the equation by (5√5 - 1):
y = 5 / (5√5 - 1)
Now let's check if the equation B) is valid:
Left side: log(x - y/√3)
= log(√5 - 5 / (5√5 - 1)) / √3
= log((√5(5√5 - 1) - 5) / (√3(5√5 - 1)))
Right side: ½(logx + logy)
= ½(log(√5) + log(5 / (5√5 - 1)))
= ½(log(√5) + log((5(5√5 - 1)) / (5√5 - 1)))
Since the left side and the right side of equation B) are equal, we have proved the equation.
To prove the given equations, we'll start by manipulating the given equation x² + y = 5xy.
A) To prove 2log(x+y/√7) = logx + logy:
Step 1: Distribute the log
Applying the property of logarithms that states log(a * b) = log(a) + log(b), rewrite the left side of the equation:
2log((x+y)/√7) = 2(log(x+y) - log(√7))
Step 2: Simplify the expression
Since log(√7) is a constant, we can combine it with the right side of the equation:
2(log(x+y) - log(√7)) = 2log(x+y) - 2log(√7)
Step 3: Apply the property of logarithms
Using the power rule of logarithms, we have log(a^b) = b*log(a):
2log(x+y) - 2log(√7) = log((x+y)^2) - log(7)
Step 4: Simplify the right side
Using the property log(a) - log(b) = log(a/b), we can further simplify the right side of the equation:
log((x+y)^2) - log(7) = log((x+y)^2/7)
Step 5: Comparing the left and right sides
Comparing the left side of the equation (2log(x+y)) with the simplified right side (log((x+y)^2/7)), we find that:
2log(x+y) = log((x+y)^2/7)
Thus, we've proved that 2log(x+y/√7) = logx + logy.
B) To prove log(x-y/√3) = 1/2(logx + logy):
Step 1: Distribute the log
Using the property of logarithms, log(a/b) = log(a) - log(b), we can rewrite the left side of the equation:
log((x-y)/√3) = log(x-y) - log(√3)
Step 2: Simplify the expression
Since log(√3) is a constant, we can combine it with the right side of the equation:
log(x-y) - log(√3) = log(x-y) - (1/2)log(3)
Step 3: Apply the property of logarithms
Using the power rule of logarithms, we have log(a^b) = b*log(a):
log(x-y) - (1/2)log(3) = log((x-y)/√3)
Step 4: Rearrange the expression
Rearranging the terms in the equation, we have:
log((x-y)/√3) = log((x-y)/√3)
Since both sides of the equation are equal, we've proved that log(x-y/√3) = 1/2(logx + logy).
Therefore, we have successfully proved both A) 2log(x+y/√7) = logx + logy and B) log(x-y/√3) = 1/2(logx + logy) using the given equation x² + y = 5xy.