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November 7th, 2014, 03:59 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(35x^3))/(log(5x)) = 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, 10:32 PM  #2 
Senior Member Joined: Jan 2012 From: Erewhon Posts: 245 Thanks: 112  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, 10:38 PM  #3  
Senior Member Joined: Jan 2012 From: Erewhon Posts: 245 Thanks: 112  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:
CB  
November 8th, 2014, 07:37 AM  #4  
Senior Member Joined: Apr 2014 From: Europa Posts: 575 Thanks: 176  Quote:
$\displaystyle a^3+3ab(a+b+3)+b^3 = 10ab \\\;\\ a^3+3a^2b+3ab^2+9ab+b^3 = 10ab_{9ab} \\\;\\ a^3+3a^2b+3ab^2+b^3 = ab \\\;\\ (a+b)^3 = ab \\\;\\ \ log(a+b)^3=\log ab \\\;\\ 3log(a+b) = \log ab \\\;\\ 3\log(a+b) = \log\ a + \log b \\\;\\ \log(a+b) = \dfrac{1}{3}(\log a + \log b)$  

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