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If = log(

3 +
3 −
2 −
2) then Prove that (

+

)
2
=
−4
(+)
2
.

1 Answer

1 vote

Given: $f(x) = \log(3x + 3 - 2x - 2)$

We need to prove that: $\frac{(a+b)^2}{f(a)+f(b)} = -4$

First, we need to simplify $f(x)$:

$f(x) = \log(x+1)$ (by simplifying the expression inside the log)

Now, we can rewrite the expression we need to prove as:

$\frac{(a+b)^2}{\log(a+1) + \log(b+1)} = -4$

Using the property of logarithms, we can rewrite this as:

$\log((a+1)(b+1)) = \log(e^{-4})$

Simplifying further:

$(a+1)(b+1) = e^{-4}$

Expanding the left side and simplifying:

$ab + a + b + 1 = e^{-4}$

Now, we can use the identity $(a+b)^2 = a^2 + 2ab + b^2$:

$(a+b)^2 = a^2 + 2ab + b^2$

Dividing both sides by $f(a)+f(b)$, which is $\log(a+1) + \log(b+1)$:

$\frac{(a+b)^2}{\log(a+1) + \log(b+1)} = \frac{a^2 + 2ab + b^2}{\log(a+1) + \log(b+1)}$

Using the property of logarithms again, we can simplify the denominator:

$\frac{(a+b)^2}{\log((a+1)(b+1))} = \frac{a^2 + 2ab + b^2}{\log((a+1)(b+1))}$

Substituting the value we derived earlier for $(a+1)(b+1)$:

$\frac{(a+b)^2}{\log(e^{-4})} = \frac{a^2 + 2ab + b^2}{\log(e^{-4})}$

Simplifying:

$(a+b)^2 = a^2 + 2ab + b^2$

$4ab = -4$

Dividing both sides by 4:

$ab = -1$

Therefore, we have shown that $\frac{(a+b)^2}{f(a)+f(b)} = -4$ if $f(x) = \log(3x+3-2x-2)$.

User Andrea Black
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