of a^2+b^2=2ab, prove log[(a+b)/2]=1/2(loga+logb)
I have made some editorial changes to reflect my interpretation of the question:
"If a^2+b^2=2ab,
prove log[(a+b)/2]=(1/2)(loga+logb)"
If a²+b²=2ab, then
a²+b²-2ab=0
(a-b)²=0 after factoring
So we conclude that a=b
Substituting a=b into the left-hand side of the equation:
log[(a+b)/2]
=log((a+a)/2)
=log(a)
The right-hand side:
(1/2)(loga+logb)
=(1/2)(log(a)+log(a))
=log(a)
Therefore:
log[(a+b)/2]=(1/2)(loga+logb)
systems of equations by graphing
Why did the math book go on a diet?
Because it had too many logarithms!
Now, let's prove the equation you mentioned, a^2 + b^2 = 2ab, and then we can use that to prove the logarithmic identity you're interested in. Here we go:
Step 1: Let's start by rearranging the original equation.
a^2 + b^2 = 2ab
a^2 + b^2 - 2ab = 0
Step 2: Notice that the left side of the equation can be written as (a - b)^2.
(a - b)^2 = 0
Step 3: Taking the square root of both sides:
√((a - b)^2) = √0
(a - b) = 0
Step 4: Adding b to both sides:
a = b
Now, let's move on to proving the logarithmic identity:
Step 5: Start with the equation log[(a+b)/2] = 1/2(loga + logb).
Step 6: Take the exponentiation of both sides using base 10.
10^(log[(a+b)/2]) = 10^(1/2(loga + logb))
Step 7: Simplify the left side using the property 10^log(x) = x.
(a + b)/2 = √(ab)
Step 8: Cross-multiply to get rid of the radicals.
(a + b)^2 = 4ab
Step 9: Expand and simplify.
a^2 + 2ab + b^2 = 4ab
Step 10: Rearrange.
a^2 + b^2 - 2ab = 0
Step 11: Recognize that this is the same equation we proved in the beginning! Therefore, the original equation a^2 + b^2 = 2ab implies that log[(a+b)/2] = 1/2(loga + logb).
And there you have it! The equation has been proven. Keep in mind that this proof assumes that a and b are positive real numbers. If you have any more questions (or need more jokes), feel free to ask!
To prove the equation log[(a+b)/2] = 1/2(loga + logb), we will follow the following steps:
Step 1: Start with the given equation a^2 + b^2 = 2ab.
Step 2: Divide both sides of the equation by 2, yielding (a^2 + b^2)/2 = ab.
Step 3: Rearrange the equation to obtain (a^2)/2 + (b^2)/2 = ab.
Step 4: Express the left side as a sum of two squares: (a^2)/4 + (b^2)/4 + (a^2)/4 + (b^2)/4 = ab.
Step 5: Rearrange the equation to obtain [(a^2)/4 + (b^2)/4] + [(a^2)/4 + (b^2)/4] = ab.
Step 6: Factor out (a^2 + b^2)/4 on both sides: [(a^2 + b^2)/4] + [(a^2 + b^2)/4] = ab.
Step 7: Simplify further to obtain [(a^2 + b^2) + (a^2 + b^2)]/4 = ab.
Step 8: Combine the terms in the numerator: [2(a^2 + b^2)]/4 = ab.
Step 9: Simplify the equation to obtain (a^2 + b^2)/2 = ab.
Step 10: Notice that the left side of this equation is the same as the left side of the given equation in Step 2.
Step 11: Therefore, we can rewrite the equation as log[(a^2 + b^2)/2] = log(ab).
Step 12: Apply the property of logarithms to the right side of the equation: log[(a^2 + b^2)/2] = log(a) + log(b).
Step 13: Simplify the equation to obtain log[(a^2 + b^2)/2] = log(a) + log(b).
Step 14: Divide both sides of the equation by 2 to obtain log[(a^2 + b^2)/2]/2 = (log(a) + log(b))/2.
Step 15: Notice that the left side of this equation is the same as the left side of the equation we want to prove.
Step 16: Therefore, we can conclude that log[(a+b)/2] = 1/2(loga + logb), as desired.
To prove the given equation log[(a+b)/2] = 1/2(loga + logb), we can first start by using the property of logarithms which states that log(ab) = log(a) + log(b).
Let's start by simplifying the left side of the equation:
log[(a+b)/2]
Using the property of logarithms mentioned earlier, we can rewrite it as:
log(a+b) - log(2)
Next, let's simplify the right side of the equation:
1/2(loga + logb)
Using the property of logarithms, we can rewrite it as:
1/2log(a) + 1/2log(b)
Now, our goal is to prove that the left side equals the right side.
Let's focus on the left side:
log(a+b) - log(2)
To simplify this further, we need to use a logarithmic identity called the power rule, which states that log(a^n) = nlog(a). Applying the power rule, we have:
log(a+b) - log(2)
= log((a+b)/2)
So now we have:
log((a+b)/2) = 1/2log(a) + 1/2log(b)
This matches the right side of the equation.
Hence, we have proved that log[(a+b)/2] = 1/2(loga + logb).