# Logarithmic

• Jan 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
• Jan 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 $\displaystyle 6^x$
Then let $\displaystyle X=6^x$ and solve the quadratic equation ;)
• Jan 15th 2009, 12:06 PM
flyingsquirrel
Hi,
Quote:

Originally Posted by Haris

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

Is that correct?
• Jan 16th 2009, 09:21 AM
Plato
$\displaystyle \begin{gathered} 4\left( {6^{2x} } \right) - 15\left( {6^x } \right) - 4 = 0 \hfill \\ \left[ {4\left( {6^x } \right) + 1} \right]\left[ {\left( {6^x } \right) - 4} \right] = 0 \hfill \\ \left( {6^x } \right) \ne \frac{{ - 1}} {4}\,\& \,\left( {6^x } \right) = 4 \hfill \\ \end{gathered}$
Can you finish?
• Jan 16th 2009, 09:24 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
• Jan 16th 2009, 09:37 AM
Haris
x=ln(4)/ln(6) ?
• Jan 16th 2009, 09:42 AM
ursa
Quote:

x=ln(4)/ln(6) ?
yup...that's it(Wink)