Logarithmic

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January 15th 2009, 11:29 AM
Haris

Logarithmic

Solve http://ta2.maths.ed.ac.uk:8080/maple...eiccknmkbi.gif.
Could someone help explain how to solve this? Cheers
January 15th 2009, 11:39 AM
Moo

Hello,

Quote:

Originally Posted by Haris

Solve http://ta2.maths.ed.ac.uk:8080/maple...eiccknmkbi.gif.
Could someone help explain how to solve this? Cheers

Multiply both sides by $6^x$
Then let $X=6^x$ and solve the quadratic equation ;)
January 15th 2009, 12:06 PM
flyingsquirrel

Hi,

Quote:

Originally Posted by Haris

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

If you are supposed to know what $\sinh$ is you can claim that

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

And this equation can easily be solved using the inverse function of $\sinh$ .
January 16th 2009, 08:40 AM
Haris

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

Is that correct?
January 16th 2009, 09:21 AM
Plato

Follow Moo's advice.
$\begin{gathered}<br /> 4\left( {6^{2x} } \right) - 15\left( {6^x } \right) - 4 = 0 \hfill \\<br /> \left[ {4\left( {6^x } \right) + 1} \right]\left[ {\left( {6^x } \right) - 4} \right] = 0 \hfill \\ \left( {6^x } \right) \ne \frac{{ - 1}}<br /> {4}\,\& \,\left( {6^x } \right) = 4 \hfill \\ <br /> \end{gathered}$
Can you finish?
January 16th 2009, 09:24 AM
ursa

Quote:

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

Is that correct?

this is not the right answer
check again
January 16th 2009, 09:37 AM
Haris

x=ln(4)/ln(6) ?
January 16th 2009, 09:42 AM
ursa

Quote:

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

yup...that's it(Wink)

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