# Express the logarithm in terms a,b,c

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.

-
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

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

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, $\log_{10}(100)=\log_{10}(10^2) = 2$.

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

-
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

1. $\quad-a$
2. $\quad5a-5b -c$
3. $\quad2+ 2a -4b -\tfrac{c}{3}$