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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 $a$,$b$,$c$.

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


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

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

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Didn't the answers to one of your previous questions help? They apply to this set, too. – J. M. Nov 29 '11 at 14:22
I will investigate, thanks. – erimar77 Nov 29 '11 at 14:29
up vote 2 down vote accepted

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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Regarding the third, I'm ending up with 2log_10(100A) - 4log_10 B + (1/3)log_10(C). Where am I going wrong with the first part? I'm thinking something needs to be done with the 100, perhaps take the square root when I'm applying the properties of logs? – erimar77 Nov 29 '11 at 19:27
You're close. You've missed the sign on the "1/3..." term. $100=10^2$ so $\log_{10}(100)=\log_{10}(10^2)=2$ :) – Bill Cook Nov 29 '11 at 19:44
AH! I see now that you can further expand on 100A by separating them out into an addition. – erimar77 Nov 29 '11 at 19:54
You've got it. :) – Bill Cook Nov 29 '11 at 19:55

The solutions (for your reference)

  1. $\quad-a$

  2. $\quad5a-5b -c$

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

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Rameez, for future reference, on this site we like explanations to accompany solutions. Bill Cook's answer above does this nicely (and leaves the task of solving the second and third problems to the user who asked the question, which is a good pedagogical practice.) – Zev Chonoles Nov 29 '11 at 16:43
The label "answers" is more appropriate than "solutions." There is all too often confusion between the two notions. The answers may very well be useful to the OP as a check. – André Nicolas Nov 29 '11 at 18:25
Oh alright. Sorry! :) – Rameez Kakodker Nov 30 '11 at 6:06

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