Solve http://ta2.maths.ed.ac.uk:8080/maple...eiccknmkbi.gif.

Could someone help explain how to solve this? Cheers

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- Jan 15th 2009, 11:29 AMHarisLogarithmic
Solve http://ta2.maths.ed.ac.uk:8080/maple...eiccknmkbi.gif.

Could someone help explain how to solve this? Cheers - Jan 15th 2009, 11:39 AMMoo
- Jan 15th 2009, 12:06 PMflyingsquirrel
Hi,If you are supposed to know what $\displaystyle \sinh$ is you can claim that

$\displaystyle 6^x-6^{-x}=\frac{15}{4} \Longleftrightarrow \mathrm{e}^{x\ln 6}-\mathrm{e}^{-x\ln 6}=\frac{15}{4} \Longleftrightarrow \sinh(x\ln6)=\frac{15}{8}$.

And this equation can easily be solved using the inverse function of $\displaystyle \sinh$. - Jan 16th 2009, 08:40 AMHaris
I got x=1+sqrt(20)/2 and x=1-sqrt(20)/2

Is that correct? - Jan 16th 2009, 09:21 AMPlato
Follow Moo's advice.

$\displaystyle \begin{gathered}

4\left( {6^{2x} } \right) - 15\left( {6^x } \right) - 4 = 0 \hfill \\

\left[ {4\left( {6^x } \right) + 1} \right]\left[ {\left( {6^x } \right) - 4} \right] = 0 \hfill \\ \left( {6^x } \right) \ne \frac{{ - 1}}

{4}\,\& \,\left( {6^x } \right) = 4 \hfill \\

\end{gathered} $

Can you finish? - Jan 16th 2009, 09:24 AMursaQuote:

I got x=1+sqrt(20)/2 and x=1-sqrt(20)/2

Is that correct?

this is not the right answer

check again - Jan 16th 2009, 09:37 AMHaris
x=ln(4)/ln(6) ?

- Jan 16th 2009, 09:42 AMursaQuote:

x=ln(4)/ln(6) ?