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1. Let \(a_0=-2,b_0=1\), and for \(n\geq 0, let \begin{align*}a_{n+1}&=a_n+b_n+\sqrt{a_n^2+b_n^2},\\b_{n+1}&=a_n+b_n-\sqrt{a_n^2+b_n^2}.\end{align*}\)Find\( \frac{1}{a_{2012}} + \frac{1}{b_{2012}}.\)

 

2. Suppose that \((\log a/b)^2 + (\log ab)^2 = 20\) and \((\log a - \log b) \log ab = 8.\) Find \($|\log a |$.\)

 

3. Find all the solutions to \(\sqrt[3]{\sqrt[3]{5 \sqrt{2} + x} + \sqrt[3]{5 \sqrt{2} - x}} = \sqrt{2}.\)Enter all the solutions, separated by commas.

 

4. Is \(f(x) = \log (x + \sqrt{1 + x^2})\)an even function, odd function, or neither? Enter "odd", "even", or "neither".

 Jun 22, 2019
 #1
avatar+103069 
+1

2.     [  log (a / b) ]^2  +  [ (log ab) ] ^2  =  20         and    (log a - log b) log(ab)  = 8

 

Note that

log (a / b)  = log a - log b

log (ab)  =   log a + log b

 

So we have

 

(log a - log b)^2  + (log a + log b)^2  =  20         and   (log a - log b) ( log a + log b )  = 8

 

Let log a  = u       Let log b = v

 

So

 

(u - v)^2   + ( u + v)^2  = 20                                 and        (u - v) (u + v)  = 8

u^2  - 2uv + v^2   + u^2 + 2uv + v^2  = 20                         u^2  - v^2  = 8     (2)

2u^2 + 2v^2  = 20

u^2 + v^2  = 10    (1)

 

 

Add (1)  and (2)    and we have that

 

2u^2  = 18

u^2  = 9

u = ±3  

 

So 

 

u =  log  a  = ± 3

 

So

 

l log a  l     =   l ± 3 l  =      3

 

 

cool cool cool

 Jun 22, 2019
 #5
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0

Thank you so much, I appreciate your help!

Guest Jun 23, 2019
 #2
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3 -  

Solve for x:
((5 sqrt(2) - x)^(1/3) + (x + 5 sqrt(2))^(1/3))^(1/3) = sqrt(2)

Raise both sides to the power of three:
(5 sqrt(2) - x)^(1/3) + (x + 5 sqrt(2))^(1/3) = 2 sqrt(2)

Subtract (x + 5 sqrt(2))^(1/3) from both sides:
(5 sqrt(2) - x)^(1/3) = 2 sqrt(2) - (x + 5 sqrt(2))^(1/3)

 

Raise both sides to the power of three:
5 sqrt(2) - x = (2 sqrt(2) - (x + 5 sqrt(2))^(1/3))^3

Subtract (2 sqrt(2) - (x + 5 sqrt(2))^(1/3))^3 from both sides:
5 sqrt(2) - x - (2 sqrt(2) - (x + 5 sqrt(2))^(1/3))^3 = 0

5 sqrt(2) - x - (2 sqrt(2) - (x + 5 sqrt(2))^(1/3))^3 = -6 sqrt(2) + 24 (x + 5 sqrt(2))^(1/3) - 6 sqrt(2) (x + 5 sqrt(2))^(2/3):
-6 sqrt(2) + 24 (x + 5 sqrt(2))^(1/3) - 6 sqrt(2) (x + 5 sqrt(2))^(2/3) = 0


Simplify and substitute y = (x + 5 sqrt(2))^(1/3).
-6 sqrt(2) + 24 (x + 5 sqrt(2))^(1/3) - 6 sqrt(2) (x + 5 sqrt(2))^(2/3) = -6 sqrt(2) + 24 (x + 5 sqrt(2))^(1/3) - 6 sqrt(2) ((x + 5 sqrt(2))^(1/3))^2
 = -6 sqrt(2) y^2 + 24 y - 6 sqrt(2):
-6 sqrt(2) y^2 + 24 y - 6 sqrt(2) = 0

Divide both sides by -6 sqrt(2):
y^2 - 2 sqrt(2) y + 1 = 0

 

Subtract 1 from both sides:
y^2 - 2 sqrt(2) y = -1

Add 2 to both sides:
y^2 - 2 sqrt(2) y + 2 = 1


Write the left hand side as a square:
(y - sqrt(2))^2 = 1

Take the square root of both sides:
y - sqrt(2) = 1 or y - sqrt(2) = -1

Add sqrt(2) to both sides:
y = 1 + sqrt(2) or y - sqrt(2) = -1

 

Substitute back for y = (x + 5 sqrt(2))^(1/3):
(x + 5 sqrt(2))^(1/3) = 1 + sqrt(2) or y - sqrt(2) = -1

Raise both sides to the power of three:
x + 5 sqrt(2) = (1 + sqrt(2))^3 or y - sqrt(2) = -1


Subtract 5 sqrt(2) from both sides:
x = (sqrt(2) + 1)^3 - 5 sqrt(2) or y - sqrt(2) = -1

(sqrt(2) + 1)^3 - 5 sqrt(2) = 7:
x = 7 or y - sqrt(2) = -1

Add sqrt(2) to both sides:
x = 7 or y = sqrt(2) - 1

 

Substitute back for y = (x + 5 sqrt(2))^(1/3):
x = 7 or (x + 5 sqrt(2))^(1/3) = sqrt(2) - 1

Raise both sides to the power of three:
x = 7 or x + 5 sqrt(2) = (sqrt(2) - 1)^3

 

Subtract 5 sqrt(2) from both sides:
x = 7 or x = (sqrt(2) - 1)^3 - 5 sqrt(2)
sqrt(2) - 1)^3 - 5 sqrt(2) = -7:

 

x = 7              or               x = -7

 Jun 22, 2019
 #7
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0

Thank you for your work!

Guest Jun 23, 2019
 #3
avatar+7711 
+1

4)

\(f(x) = \log \left(x + \sqrt{1 +x^2}\right)\\ \begin{array}{rcll} f(-x) &=& \log \left(-x + \sqrt{1 + x^2}\right)\\ &=&\log \left(\dfrac{(-x+\sqrt{1+x^2})(x + \sqrt{1+x^2})}{x+\sqrt{1+x^2}}\right)\\ &=&\log \left(\dfrac{(\sqrt{1+x^2})^2-x^2}{x+\sqrt{1+x^2}}\right)\\ &=&\log \left(\dfrac{1}{x+\sqrt{1+x^2}}\right)\\ &=&\log \left(\left(x + \sqrt{1+x^2}\right)^{-1}\right)\\ &=&-\log\left(x+\sqrt{1+x^2}\right)\\ &=&-f(x)\\ \end{array}\\ \therefore \text{It is an odd function.}\)

.
 Jun 22, 2019
 #8
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0

Thank you very much as well!

Guest Jun 23, 2019
 #4
avatar+23048 
+3

1.
Let \(a_0=-2,b_0=1\), and for\( n\ge 0, let \begin{align*}a_{n+1}&=a_n+b_n+\sqrt{a_n^2+b_n^2},\\ b_{n+1}&=a_n+b_n-\sqrt{a_n^2+b_n^2}.\end{align*}\)
Find\( \dfrac{1}{a_{2012}} + \dfrac{1}{b_{2012}}\).

 

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{1}{a_{n+1}} + \dfrac{1}{b_{n+1}}} &=& \dfrac{1}{a_n+b_n+\sqrt{a_n^2+b_n^2}} + \dfrac{1}{a_n+b_n-\sqrt{a_n^2+b_n^2}} \\\\ &=& \dfrac{a_n+b_n-\sqrt{a_n^2+b_n^2}+a_n+b_n-\sqrt{a_n^2+b_n^2}} {\left(a_n+b_n+\sqrt{a_n^2+b_n^2}\right)\left(a_n+b_n-\sqrt{a_n^2+b_n^2}\right)} \\\\ &=& \dfrac{2(a_n+b_n)} {\left(a_n+b_n\right)^2-\left(\sqrt{a_n^2+b_n^2}\right)^2} \\\\ &=& \dfrac{2(a_n+b_n)} {a_n^2+2a_nb_n+b_n^2-(a_n^2+b_n^2)} \\\\ &=& \dfrac{2(a_n+b_n)} { 2a_nb_n} \\\\ &=& \dfrac{ a_n+b_n } { a_nb_n} \\\\ &=& \mathbf{\dfrac{1}{a_{n}} + \dfrac{1}{b_{n}}} \\ \hline \mathbf{\dfrac{1}{a_{n+1}} + \dfrac{1}{b_{n+1}}} &=& \mathbf{\dfrac{1}{a_{n}} + \dfrac{1}{b_{n}}} = \ldots = \mathbf{\dfrac{1}{a_{1}} + \dfrac{1}{b_{1}}} \\\\ \dfrac{1}{a_{1}} + \dfrac{1}{b_{1}} &=& \dfrac{1}{-1+\sqrt{5}} + \dfrac{1}{-1-\sqrt{5}} = \dfrac{1}{2} \\\\ \hline \mathbf{\dfrac{1}{a_{2012}}+\dfrac{1}{b_{2012}}} &=& \dfrac{1}{a_{1}} + \dfrac{1}{b_{1}}\mathbf{ =\dfrac{1}{2}} \\ \hline \end{array}\)

 

laugh

 Jun 22, 2019
 #6
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+1

Thank you for you thorough steps, I really understand now!

Guest Jun 23, 2019

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