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Expert: - 3/12/2009


Question

Questions 4 - 12
I am really stuck on questions 9, 10, 11 and 12. Can you please help?

Answer

Listing
Example 11:
a. If  A = log4(3), then 4^A = 3.
  If  B = log2(sqrt(3)) then:
      B = 1/2 log2(3) and
     2B = log2(3), then
      2^(2B) = 3
So  2^(2B) = 4^A
2^(2B) = (2^2)^A

2^(2B) = 2^(2A)

2B = 2A,  so A = B
.....................
b) 2 log2(y) = log4(3) + log2(x)

    3^y = 9^x  -->  
    3^y = (3^2)^x  -->
    3^y = 3^2x -->  
    y = 2x.

2 log2(y) = log2(sqrt(3)) + log2(x)  << This is henceing.
(whatever that means)

2 log2(2x) = log2(sqrt(3)) + log2(x)

2 (1 + log2(x)) = log2(sqrt(3)) + log2(x)
2 + 2log2(x) = log2(sqrt(3)) + log2(x)
log2(x) = log2(sqrt(3)) - 2

log2(x) = log2(sqrt(3)/4)

x = sqrt(3)/4
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Example 10:

a^x = b^y = (ab)^xy

(ab)^xy = a^xy b^xy =

(a^x)^y (b^y)^x =

(b^y)^y (b^y)^x =

b^(y^2) b^(xy) =

b^(y^2 + xy) = b^y,  ==>

y^2 + xy = y

y^2 + xy - y = 0
y(y + x - 1) = 0

So either y = 0 or

x + y = 1.

If y = 0, then:

a^x = b^0 = (ab)^0

a^x = 1, then  x = 0.  Hmmmmm.
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Example 9:

xy = 64   and  logx(y) + logy(x) = 5/2.

Basic log property: logx(y) = 1/logy(x).  

Call logx(y) = A for convenience.  Then:

A + 1/A = 5/2

2A^2 + 2 = 5A
2A^2 - 5A + 2 = 0

(2A - 1)(A - 2) = 0

A = 1/2, or A = 2.

So  logx(y) = 2.  (never mind the 1/2 -- this example is symmetric in x and y)

So y = x^2

Now  xy = 64, so x^3 = 64 and  x = 4, so y = 16.  You can finish.
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Paul Klarreich

Expertise

I can answer questions in basic to advanced algebra (theory of equations, complex numbers), precalculus (functions, graphs, exponential, logarithmic, and trigonometric functions and identities), basic probability, and finite mathematics, including mathematical induction. I can also try (but not guarantee) to answer questions on Abstract Algebra -- groups, rings, etc. and Analysis -- sequences, limits, continuity. I won't understand specialized engineering or business jargon.

Experience

I taught at a two-year college for 25 years, including all subjects from algebra to third-semester calculus.

Education/Credentials
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