Totally lost

**macca101** · April 17th 2006, 11:12 AM

By solving a suitable equation in k, find a positive integer k such that...

$ (x^2)^{20-k}(\frac{1}{x})^k=x $

**TD!** · April 17th 2006, 11:24 AM

Apply properties of exponents:

$ \begin{gathered} \left( {x^2 } \right)^{20 - k} \left( {\frac{1} {x}} \right)^k = x \hfill \\ x^{2\left( {20 - k} \right)} \left( {x^{ - 1} } \right)^k = x \hfill \\ x^{40 - 2k} x^{ - k} = x \hfill \\ x^{40 - 3k} = x \hfill \\ \Rightarrow 40 - 3k = 1 \Leftrightarrow k = 13 \hfill \\ \end{gathered} $

**macca101** · April 17th 2006, 11:42 AM

Thanks very much

**TD!** · April 17th 2006, 11:45 AM

You're welcome, hopefully clear to you now?

**macca101** · April 17th 2006, 11:48 AM

Originally Posted by TD!

You're welcome, hopefully clear to you now?

It always is after the fact

**TD!** · April 17th 2006, 11:52 AM

Better late than never

Thread: Totally lost

LinkBack

Thread Tools

Display

Totally lost

Similar Math Help Forum Discussions

Totally lost with Sorgenfry

Factorable Trig equations (I'm totally lost with this one)

Totally Lost - Functions

Totally baffled

totally frustrated

Search Tags