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Math Help - Bit rusty on my log rearranging

  1. #1
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    Bit rusty on my log rearranging

    Can you help me with these

    Rewrite these relations in index form (that is, without using logarithims):

    (a) logα (x + y) = logα x + logα y
    (b) log ₁₀ x = 3 + log ₁₀ y
    (c) log₃ x = 4log₃y
    (d) 2log₂x + 3log₂y – 4log₂z = 0
    (e) xlogα 2 = log α y
    (f) log α x - log α y = n log α z
    (g) ˝log₂ x + = ⅓log₂ y – 1
    (h) 2 log₃(2x + 1) = 3 log₃ (2x – 1)
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Joel View Post
    Can you help me with these

    Rewrite these relations in index form (that is, without using logarithims):

    (a) logα (x + y) = logα x + logα y
    (b) log ₁₀ x = 3 + log ₁₀ y
    (c) log₃ x = 4log₃y
    (d) 2log₂x + 3log₂y – 4log₂z = 0
    (e) xlogα 2 = log α y
    (f) log α x - log α y = n log α z
    (g) ˝log₂ x + = ⅓log₂ y – 1
    (h) 2 log₃(2x + 1) = 3 log₃ (2x – 1)
    a)log_{a}(x+y) = log_{a} x + log_{a} y

    log_{a} (x+y) =  log_{a} xy  I use  log A + log B = log (AB)

     (x+y) = xy

    b) log_{10} x =3 + log_{10} y

    x=10^{3+log_{10} y }

    x= 10^3 (10^{log_{10} y })

    x= 1000(y) since  A^{log_{A} c} = c


    e)xlog_{a} 2 = log_{a} y

     log_{a}2^x= log_{a} y since (a) log c = log c^a

     2^x = y

    (g) ˝log₂ x + = ⅓log₂ y – 1

    g)1/2log_{2}x = 1/3log_{2} y -1

    log_{2} x^{1/2} - log_{2} y^{1/3} = -1

    log_{a} y^{1/3} - log_{2} x^{1/2} = 1 multiply with -1

    log_{a} \dfrac{y^{1/3}}{x^{1/2}} = 1

    \dfrac{y^{1/3}}{x^{1/2}} = a
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  3. #3
    MHF Contributor Amer's Avatar
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    some log rules I attach them because in sometimes latex it is not clear if I use powers

    Bit rusty on my log rearranging-log-rules.jpg
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  4. #4
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    Hello, Joel!

    Some of your bases didn't show up . . . I'll use b.


    Rewrite in index form (that is, without using logarithims):

    (a)\;\;\log_b(x+y) \:=\:\log_b(x) + \log_b(y)
    We have: . \log_b(x+y) \:=\:\log_b(xy)

    Then: . x+y \:=\:xy



    (b)\;\;\log_b(x) \:= \:3 + \log_b(y)
    We have: . \log_b(x) \;=\;\log_b(b^3) + \log_b(y) \;=\;\log_b(b^3y)

    Then: . x \;=\;b^3y



    (c)\;\;\log_b(x) \:=\: 4\log_b(y)
    We have: . \log_b(x) \;=\;\log_b(y^4)

    Then: . x \;=\;y^4



    (d)\;\;2\log_b(x) + 3\log_b(y) - 4\log_b(z) \:=\:0
    We have: . \log_b(x^2) + \log_b(y^3) - \log_b(z^4) \:=\:0 \quad\Rightarrow\quad \log_b\left(\frac{x^2y^3}{z^4}\right) \:=\:0

    Then: . \frac{x^2y^3}{x^4} \:=\:1



    (e)\;\;x\log_b(2) \:=\: \log_b(y)
    We have: . \log_b(2^x) \:=\:\log_b(y)

    Then: . 2^x\:=\:y



    (f)\;\;\log_b(x) - \log_b(y) \:=\: n\log_b(z)

    We have: . \log_b(x) - \log_b(y) - \log_b(z^n) \:=\:0 \quad\Rightarrow\quad \log_b\left(\frac{xy}{z^n}\right) \:=\:0

    Then: . \frac{xy}{z^n} \;=\;1



    (g)\;\;\tfrac{1}{2}\log_b(x) \:=\: \tfrac{1}{2}\log_b(y) - 1

    We have: . \log_b\left(x^{\frac{1}{2}}\right) \;=\;\log_b\left(y^{\frac{1}{2}}\right) - \log_b(b)

    . . \log_b\left(\sqrt{x}\right) - \log_b\left(\sqrt{y}\right) + \log_b(b) \;=\;0 \quad\Rightarrow\quad \log_b\left(\frac{\sqrt{x}}{\sqrt{y}}\cdot b\right) \;=\;0

    Then: . \frac{b\sqrt{x}}{\sqrt{y}} \:=\:1



    (h)\;\;2\log_b(2x + 1) \:=\: 3\log_b (2x – 1)

    We have: . \log_b(2x+1)^2 \;=\;\log_b(2x-1)^3

    Then: . (2x+1)^2 \;=\;(2x-1)^3

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