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Math Help - Logs

  1. #1
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    Exclamation Logs

    Suppose that f(x) = log (base 4) x

    1. If f(3) = 0.79 and f(5) = 1.16, then evaluate f(75)

    2. If f(3) is 0.79 find log (base 8) 3

    3. Show that f(x) + f(1/x) = 0

    THANKS GUYS!
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  2. #2
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    Quote Originally Posted by goliath View Post
    Suppose that f(x) = log (base 4) x

    1. If f(3) = 0.79 and f(5) = 1.16, then evaluate f(75)

    2. If f(3) is 0.79 find log (base 8) 3

    3. Show that f(x) + f(1/x) = 0

    THANKS GUYS!
    1.  f(75)=log_{4}75=log_{4}(5^2\times 3)=log_{4}(5^2)+log_{4} 3
    =2log_{4}5+log_{4}3=2f(5)+f(3)
    now put values of f(5) and f(3)
    2.
     log_{8} 3 =\frac{log_{2} {3}}{log_{2}{8}}=\frac{log_{2} {3}}{3log_{2}{2}} \ now \ continue
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  3. #3
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    Quote Originally Posted by goliath View Post
    Suppose that f(x) = log (base 4) x

    1. If f(3) = 0.79 and f(5) = 1.16, then evaluate f(75)

    2. If f(3) is 0.79 find log (base 8) 3

    3. Show that f(x) + f(1/x) = 0

    THANKS GUYS!
    You're expecetd to apply the various log rules you've been taught.

    1. Note that 75 = 3 \times 5^2.

    ----------------------------------------------------------------------------

    2. \log_8 3 = x \Rightarrow 8^x = 3 \Rightarrow (2^3)^x = 3 \Rightarrow 2^{3x} = 3 \Rightarrow \log_2 3 = 3x .... (1)

    \log_4 3 = 0.79 \Rightarrow 4^{0.79} = 3 \Rightarrow (2^2)^{0.79} = 3 \Rightarrow 2^{1.58} = 3 \Rightarrow \log_2 3 = 1.58 .... (2)

    Substitute (2) into (1) and solve for x.

    There are undoubtedly easier ways of doing this but they're my first thoughts.

    -------------------------------------------------------------------------------

    3. Note that f\left(\frac{1}{x}\right) = f(x^{-1}) = \log_4 x^{-1}.
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  4. #4
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    2.another method
    log_{8} 3=\frac{log_{4} 3}{log_{4}8}
    =\frac{f(3)}{log_{4}{4} \times 2}
    =\frac{f(3)}{(log_{4}{4}) + (log_{4}{ 2})}
    =\frac{f(3)}{1+\frac{1}{log_{2}{4}}}
    =\frac{f(3)}{1+\frac{1}{2}}
    continue...
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    You're expecetd to apply the various log rules you've been taught.

    1. Note that 75 = 3 \times 5^2.

    ----------------------------------------------------------------------------

    2. \log_8 3 = x \Rightarrow 8^x = 3 \Rightarrow (2^3)^x = 3 \Rightarrow 2^{3x} = 3 \Rightarrow \log_2 3 = 3x .... (1)

    \log_4 3 = 0.79 \Rightarrow 4^{0.79} = 3 \Rightarrow (2^2)^{0.79} = 3 \Rightarrow 2^{1.58} = 3 \Rightarrow \log_2 3 = 1.58 .... (2)

    Substitute (2) into (1) and solve for x.

    There are undoubtedly easier ways of doing this but they're my first thoughts.

    -------------------------------------------------------------------------------

    3. Note that f\left(\frac{1}{x}\right) = f(x^{-1}) = \log_4 x^{-1}.

    Ok, for 3, where would I go after the inverse. I think it would be log_4 x + log_4 x^-1, but then what? Could you please explain? Thank you
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  6. #6
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    Quote Originally Posted by goliath View Post
    Ok, for 3, where would I go after the inverse. I think it would be log_4 x + log_4 x^-1, but then what? Could you please explain? Thank you
    You're expected to know and be able to apply the log rule \log_a b^m = m \log_a b.
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    You're expected to know and be able to apply the log rule \log_a b^m = m \log_a b.

    sweet, I got it. Now I have 1 more question

    4. Find a function g such that f(g(x)) = 10f(x).
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  8. #8
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    Quote Originally Posted by goliath View Post
    sweet, I got it. Now I have 1 more question

    4. Find a function g such that f(g(x)) = 10f(x).
    \log_4 g(x) = 10 \log_4 x. Apply the rule I reminded you of and then draw a conclusion after comparing each side.
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  9. #9
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    could I use g(x) = x^10 and then put that in as x and then put the 10 to the front of the log?
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  10. #10
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    Quote Originally Posted by goliath View Post
    could I use g(x) = x^10 and then put that in as x and then put the 10 to the front of the log?
    g(x) = 10^x is the correct answer. The End.
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  11. #11
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    logs

    originaly posted by goliath
    given log base 4 3=.79 find log base 8 3. All parts answered Iadd a simplication for the latter calculation

    log ( base 8) 8=1 log (base4 ) 8=3/2
    therefore 3/2 is the factor to go from base 8 to base 4. going from base 4 to base 8 you multiply by 2/3

    2/3log(base4)3=log (base8)3=2/3(,79)=.527


    bjh
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