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 November 1st, 2015, 02:26 PM #1 Newbie   Joined: Jan 2015 From: Philadelphia Posts: 20 Thanks: 2 Expanding and simplifying logarithms So there is no way to know how to do this unless you know some complicated algebraic rules. For example, square root is somehow the same thing as raised to the power of 1/2. To get rid of a denominator, change the sign of the exponent and multiply it to the numerator. Bring down the exponent in front of the LN or LOG. Log mn is equal to log m- log n. Log(1) =0. Despite knowing all these things, I have no idea how to do expand and simplify this equation. Do I need to know more? (Goodness). So I'm stuck with $\displaystyle \log square root of x+1/x^2+1$ So what I did was change that square root to x+1^1/2. Then I brought the denominator up by subtracting it to the numerator. So now I have $\displaystyle \log(x+1)^{1/2}-x^2+1$. I'm wrong by the way. I don't know why. Help please. Last edited by skipjack; November 1st, 2015 at 05:25 PM. November 1st, 2015, 03:03 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 What you have posted is $\log{\sqrt{x}} + \dfrac{1}{x^2} +1$ Is that what you meant? November 1st, 2015, 05:33 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2201 $\displaystyle \log\frac{\sqrt{x+1}}{x^2+1} = \log(x+1)^{1/2} + \log\left(x^2+1\right)^{-1} = \frac12\log(x + 1) - \log\left(x^2+1\right)$ Tags expanding, logarithms, simplifying 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 mschoer Algebra 1 April 7th, 2014 08:27 AM rgarcia128 Calculus 5 September 27th, 2011 10:59 PM jaredbeach Algebra 4 September 2nd, 2011 04:52 PM everettjsj2 Calculus 2 February 28th, 2010 01:49 PM jaredbeach Calculus 4 December 31st, 1969 04:00 PM

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