View Full Version : Expand x^3(x-5)^6 over square root of x-4
BigTreeWoody
Jun 3, 2009, 07:18 PM
Expand x^3(x-5)^6 over square root of x-4
Evaluate log[base 6]square root of 6
Perito
Jun 3, 2009, 07:28 PM
Expand \frac {x^3(x-5)^6}{\sqrt {x^{-4}}
Evaluate \Large log_6 \left( \sqrt {6} \right)
Expand \Large \frac {x^3(x-5)^6}{\sqrt {x^{-4}}
\sqrt{x^{-4}}=\sqrt {\frac {1}{x4} }= \frac {1}{x^2}
You should be able to take it from there.
\Large log_6 \left( \sqrt {6} \right)=log_6 \left( 6^{\frac 12} \right)
Now, ask yourself, what does log_6(x) mean? It means the power that 6 has to be raised to in order to get x. So, what power does 6 have to be raised to in order to get 6^{\frac 12} (hint: don't think too hard.)?
decent
Jun 7, 2009, 12:06 PM
Expand \Large \frac {x^3(x-5)^6}{\sqrt {x^{-4}}
\sqrt{x^{-4}}=\sqrt {\frac {1}{x4} }= \frac {1}{x^2}
You should be able to take it from there.
\Large log_6 \left( \sqrt {6} \right)=log_6 \left( 6^{\frac 12} \right)
Now, ask yourself, what does log_6(x) mean? It means the power that 6 has to be raised to in order to get x. So, what power does 6 have to be raised to in order to get 6^{\frac 12} (hint: don't think too hard.)?
So what are the answers, Perito?
Unknown008
Jun 8, 2009, 09:54 AM
Hey decent, we're not supposed to directly give the answers here. What if the OP gets the answer right, but has learned nothing? He/She'll not be able to do such questions the day of his/her exams! :mad: I'll PM you the answers if you want to know the solution.
decent
Jun 10, 2009, 09:46 AM
Hey decent, we're not supposed to directly give the answers here. What if the OP gets the answer right, but has learned nothing? He/She'll not be able to do such questions the day of his/her exams! :mad: I'll PM you the answers if you want to know the solution.
Yeah you may send a PM to me! Thanks a lot..! :)
Unknown008
Jun 10, 2009, 09:49 AM
Did you receive it?