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 November 7th, 2014, 02:59 PM #1 Newbie   Joined: Nov 2014 From: Canada Posts: 1 Thanks: 0 Need Help with Homework (Logarithms) Hello MathForum, I'm not sure if this is the right place to ask for homework questions, but I'm in desperate need of help right now. I have absolutely no clue or understand how to do these questions, so I hope you can link/show me some tutorials or videos specifically for these questions. 1. Express in exponential form b) -3log_5(z) - x^2 = 2x + 1 8. Solve (log(35-x^3))/(log(5-x)) = 3. 9. If a^3+3ab(a+b+3)+b^3=10ab, show that log(a+b) = 1/3(log(a)+log (b)). 7. In a particular circuit, the current, I, in amperes can be found using the formula: I(t) = 1.251 - 1.251(10^-0.02668t), after t seconds. a) what is the current in the circuit after 1 minute? b) rewrite the equation for I(t), solving for t c) use the equation in part b) to determine the time it takes for the current to reach 0.5A.
November 7th, 2014, 09:32 PM   #2
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Joined: Jan 2012
From: Erewhon

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Quote:
 Originally Posted by Pancho 1. Express in exponential form b) -3log_5(z) - x^2 = 2x + 1
Rearrange so that you have the log term on one side and everythng else on the other:
$$-3\log_5(z)=x^2+2x+1 \\ \phantom{-3\log_5(z)}=(x+1)^2$$

Now use the laws of logarithms on the left:
$$-3\log_5(z)=\log_5(1/z^3) =(x+1)^2$$

Now $\log_A(b)=c$ means: $A^c=b$, so applying this to the above gives:
$$5^{(x+1)^2}=\frac{1}{z^3}$$

and I leave any further simplification to you.

CB

November 7th, 2014, 09:38 PM   #3
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Joined: Jan 2012
From: Erewhon

Posts: 245
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Quote:
 Originally Posted by Pancho 8. Solve (log(35-x^3))/(log(5-x)) = 3.
Rewrite this so it is of the form $\log($something$)=\log($something_else$)$ then you may conclude that something=something_else (assuming the logs of both are defined).

Quote:
 9. If a^3+3ab(a+b+3)+b^3=10ab, show that log(a+b) = 1/3(log(a)+log (b)).
For this you need to rewrite the first equation in the form: $(a+b)^3=...$

CB

November 8th, 2014, 06:37 AM   #4
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Quote:
 Originally Posted by Pancho 9. If a^3+3ab(a+b+3)+b^3=10ab, show that log(a+b) = 1/3(log(a)+log (b))
Let be a, b >0

$\displaystyle a^3+3ab(a+b+3)+b^3 = 10ab \\\;\\ a^3+3a^2b+3ab^2+9ab+b^3 = 10ab|_{-9ab} \\\;\\ a^3+3a^2b+3ab^2+b^3 = ab \\\;\\ (a+b)^3 = ab \\\;\\ \ log(a+b)^3=\log ab \\\;\\ 3log(a+b) = \log ab \\\;\\ 3\log(a+b) = \log\ a + \log b \\\;\\ \log(a+b) = \dfrac{1}{3}(\log a + \log b)$

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