Can you solve the equation:

- $${{\mathtt{e}}}^{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}\right)}{\mathtt{\,-\,}}{\mathtt{5}} = {\mathtt{8}}$$

e being Euler's number, approximately 2.71828.

EinsteinJr
Apr 30, 2015

#1**+5 **

$${{\mathtt{e}}}^{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}\right)}{\mathtt{\,-\,}}{\mathtt{5}} = {\mathtt{8}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\frac{{ln}{\left({{\mathtt{13}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}\right)}{\mathtt{\,\times\,}}{i}\right)}}{{ln}{\left({\mathtt{e}}\right)}}}\\

{\mathtt{x}} = {\frac{{ln}{\left({\mathtt{\,-\,}}\left({{\mathtt{13}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}\right)}\right)\right)}}{{ln}{\left({\mathtt{e}}\right)}}}\\

{\mathtt{x}} = {\frac{{ln}{\left({\mathtt{\,-\,}}\left({{\mathtt{13}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}\right)}{\mathtt{\,\times\,}}{i}\right)\right)}}{{ln}{\left({\mathtt{e}}\right)}}}\\

{\mathtt{x}} = {\frac{{ln}{\left({\mathtt{13}}\right)}}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{ln}{\left({\mathtt{e}}\right)}\right)}}\\

\end{array} \right\}$$

There you go - it has got 2 real solutions and 2 imaginary solutions.

$$\\e^{4x}-5=8\\\\

e^{4x}=13\\\\

ln(e^{4x})=ln(13)\\\\

4xln(e)=ln(13)\\\\

4x=ln(13)\\\\

x=ln(13)/4$$

that is interesting, I missed out on one of the answers

Melody
Apr 30, 2015

#3**0 **

Thanks but I already ate far too much tonight.

Don't you know how to make that image smaller?

Melody
Apr 30, 2015

#4**0 **

I was actually hoping that maybe Alan could show me how to get the other answer :/

Can you please Alan ?

Melody
Apr 30, 2015

#6**+3 **

Ok well thanks for the cookie I shall put it aside for later. :)

Perhaps you could find a little gold star for next time (I should be on a diet anyway)

The huge pic is a bit distracting in the middle of a thread but it was a really nice gesture :)

Melody
Apr 30, 2015