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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

Originally Posted by BigTreeWoody

Expand $\frac {x^3(x-5)^6}{\sqrt {x^{-4}}$

Evaluate $log_6 \left( \sqrt {6} \right)$

Expand $\frac {x^3(x-5)^6}{\sqrt {x^{-4}}$

$\sqrt{x^{-4}}=\sqrt {\frac {1}{x4} }= \frac {1}{x^2}$

You should be able to take it from there.

$log_6 \left( \sqrt {6} \right)=log_6 \left( 6^{\frac 12} \right)$

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

Originally Posted by Perito

Expand $\frac {x^3(x-5)^6}{\sqrt {x^{-4}}$

$\sqrt{x^{-4}}=\sqrt {\frac {1}{x4} }= \frac {1}{x^2}$

You should be able to take it from there.

$log_6 \left( \sqrt {6} \right)=log_6 \left( 6^{\frac 12} \right)$

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

Originally Posted by Unknown008

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?