My Math Forum Need help to solve few problems :(

 Calculus Calculus Math Forum

 August 13th, 2008, 09:06 PM #1 Newbie   Joined: Aug 2008 Posts: 4 Thanks: 0 Need help to solve few problems :( Hello all, Im new to this forum and im not sure if this is the right place to post I need help in solving the following precalculus questions I do understand some of the steps but I always get lost and cannot solve them Can someone please help me solve them? I really need to do them for tomorrow I hope I'm not asking for much here's the link to the problems: http://i187.photobucket.com/albums/x17/ ... oblems.jpg greets, amero
 August 14th, 2008, 02:25 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,644 Thanks: 2084 1.Condense the expressions to the logarithm of a single quantity: a) $\log_4{z}\,-\,\log_4{y}= \log_4{\frac{z}{y}}$ b) $\ln{x}\,-\,3\ln(x + 1)= \ln\left(\frac{x}{(x+1)^3}\right)$ 2. Simplfy the expression by removing the common factor with the smaller exponent: $2x^2(x-1)^{\frac12}-5(x-1)^{-\frac12}=(2x^3-2x^2-5)(x-1)^{-\frac12}$ 3. Perform the operations and simplify: $\frac{\frac2x\,-\,\frac{2}{x+1}}{\frac{4}{x^2\,-\,1}}\,=\,\frac12(1\,-\,\frac{1}{x})$ (or you might consider a slightly different form is "simplest") 4. Use the vertex formula to determine the vertex of the graph of the function and write the function in standard form: $f(x)\,=\,-5x^2\,-\,6x\,+\,3\,=\,-5(x\,+\,0.3)^2\,+\,3.45$ 5. Solve the equation: $(4x)^{\frac23}=(30x\,+\,4)^{\frac13}$ Cube, then factorize as (8x + 1)(2x - 4). Hence x = -1/8 or 2. 6. Since the exponent is zero, the value of the expression is 1.
 August 14th, 2008, 09:10 AM #3 Newbie   Joined: Aug 2008 Posts: 4 Thanks: 0 Re: Need help to solve few problems :( thx a lot man! now i just need 3 more that I'm also stuck on for the problem 1. a) i get the answer 12 / 5a .. is that correct? Can anyone please help http://i187.photobucket.com/albums/x17/ ... lems-1.jpg (I'm sorry that i give it in link but i have no idea how to write in LAteX ) greets and again I'm sorry if I'm asking for much :P
 August 14th, 2008, 10:54 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,644 Thanks: 2084 1. Solve each radical in simplest form: a) $\sqrt{\frac{48a}{20a^4}}\,=\,(\frac{2}{a^2})\sqrt{ \frac{3a}{5}}$ b) $\sqrt{\frac{2x}{27y}}\,=\,(\frac13)\sqrt{\frac{2x} {3y}}$ 2. Simplify by rationalizing the denominator: $\frac{\sqrt{7}}{2i\sqrt{7}}\,=\,\frac{i}{2i^2}\,=\ ,-\frac{i}{2}$ 3. Solve: $\frac{2x}{7}\,-\,\frac12\,=\,\frac{3x\,+\,1}{2}$ x = (-1/2 - 1/2)/(3/2 - 2/7) = -14/17 4. Write in standard form: $\frac{8\,-\,7i}{1\,-\,2i}\,=\,\frac{(8\,-\,7i)(1\,+\,2i)}{1-4i^2}\,=\,\frac{22\,+\,9i}{5}$ Which three of these four did you need help on?

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