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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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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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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

$\displaystyle a^3+3ab(a+b+3)+b^3 = 10ab \\\;\\ a^3+3a^2b+3ab^2+9ab+b^3 = 10ab|_{-9ab} \\\;\\ a^3+3a^2b+3ab^2+b^3 = ab \\\;\\ (a+b)^3 = ab \\\;\\ \ log(a+b)^3=\log ab \\\;\\ 3log(a+b) = \log ab \\\;\\ 3\log(a+b) = \log\ a + \log b \\\;\\ \log(a+b) = \dfrac{1}{3}(\log a + \log b)$ Tags homework, logarithms Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post ron246 Algebra 5 May 22nd, 2014 06:59 AM Mindless_1 Algebra 2 November 14th, 2012 12:02 PM hahachow Algebra 1 October 3rd, 2012 09:04 AM Cstolworthy Algebra 1 October 15th, 2008 02:25 PM Ian McPherson Calculus 2 December 31st, 1969 04:00 PM

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