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3 years ago

Log(x+3)(2x+3)+log(x+3)(x+5)=2

1 answer:

4 0

<span>log(x-3)+log(x) = 2 + log(x+2)
---------------------------
log[(x-3)x] - log(x+2) = 2
-----
log[(x^2-3x)/(x+2)] = 2
----
(x^2-3x)/(x+2) = 2^2
-------

x^2-3x = 4x+8
x^2 - 7x - 8 = 0
Factor:
(x-8)(x+1) = 0
Positive solution:
x = 8</span>

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Step-by-step explanation:

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3 years ago

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<img src="https://tex.z-dn.net/?f=%24a%2Ba%20r%2Ba%20r%5E%7B2%7D%2B%5Cldots%20%5Cinfty%3D15%24%24a%5E%7B2%7D%2B%28a%20r%29%5E%7B

riadik2000 [5.3K]

Let

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where we assume |r| < 1. Multiplying on both sides by r gives

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and subtracting this from $S_n$ gives

$(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}$

As n → ∞, the exponential term will converge to 0, and the partial sums $S_n$ will converge to

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Now, we're given

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We must have |r| < 1 since both sums converge, so

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Solving for r by substitution, we have

$\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)$

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We've already confirmed r ≠ 1, so we can simplify this to

$\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15$

It follows that

$\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12$

and so the sum we want is

$ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}$

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

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