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Suppose that:

  • $\log_{10}A = a$
  • $\log_{10}B = b$
  • $\log_{10}C = c$

I need to express the following in terms of $a$,$b$,$c$.

$\log_{10}A + 2\log_{10}(1/A)$

$\log_{10}(((AB)^5)/C)$

$\log_{10}((100A^2)/(B^4 \cdot \sqrt[3]{C}))$

Can someone give me a starting point? I will work through and post questions if I get stuck.

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    I will investigate, thanks.2011-11-29

2 Answers 2

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Remember your laws of logs:

  • $\log_{10}(XY) = \log_{10}(X)+\log_{10}(Y)$
  • $\log_{10}(X/Y) = \log_{10}(X)-\log_{10}(Y)$
  • $\log_{10}(X^Y) = Y\cdot\log_{10}(X)$

So for the first one: $\log_{10}(A)+2\cdot\log_{10}(1/A) $ $=\log_{10}(A)+2\cdot\log_{10}(A^{-1})$ $=\log_{10}(A)-2\cdot\log_{10}(A)$ $= - \log_{10}(A)=-a$

Alternatively: $\log_{10}(A)+2\cdot\log_{10}(1/A) $ $= \log_{10}(A)+2\cdot\log_{10}(1)-2\cdot\log_{10}(A) = a + 2\cdot 0 -2a = -a $ (because $\log(1)=0$)

For the third one, keep in mind $\sqrt[3]{C} = C^{1/3}$

Edit: The third...

Keep in mind the cube root is also in the denominator, so that argument is being divided as well and should have a minus sign in front of it. Next, $\log_{10}(100)=\log_{10}(10^2) = 2$.

$\log_{10}((100A^2)/(B^4 \cdot \sqrt[3]{C}))$ $=\log_{10}(100)+2\log_{10}(A)-4\log_{10}(B)-(1/3)\log_{10}(C)$ $=2+2a-4b-(1/3)c$

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    You've got it. :)2011-11-29
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The solutions (for your reference)

  1. $\quad-a$

  2. $\quad5a-5b -c$

  3. $\quad2+ 2a -4b -\tfrac{c}{3} $

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    Oh alright. Sorry! :)2011-11-30