CPhill

e^.5x=30   take the ln of both sides

ln e^.5x= ln 30   and by a log property, we can write

(.5x) lne = ln 30      remember, ln e is just 1    ... so we have....

.5x  = ln 30           

(1/2) x = ln 30     now, just multiply both sides by 2 and that will give you your answer...!!!

Nice presentation, Melody...!!!

Here's one way....

Note that  300 / 6 = 50

But 294 is one less 6 than 300

So 294 / 6   must just be 49 "6s"

Here's a slightly different approach to Melody's

Note that we can write .5  as 1/2  = 2^-1 

So we have

log₂(19)  + log_0.5 (19)   .....and...using the"change of base" rule, we can write

log(19) / log(2)  +  log(19) / log(.05) =

log(19) / log(2)  + log(19) / log(1/2) =

log(19) / log(2)  +  log(19) / log2^-1 =

log(19) / log(2)  +  log(19) / (-1) log2 =

log(19) / log(2)  +  [ (-1) log(19) /  log2 ] = 

log(19) / log(2)  -  log(19) /  log2 =   0

LOL, Alan!!!!...yeah.....Usain Bolt couldn't even beat this guy.......no matter what the distance....!!!!

And he would put the  Kenyans to shame in the marathon.....completing it in about 43 minutes....!!!

Nice answer, rosala....!!!

BTW.....where, exactly, is the "Island of Fractions" located????

Yes, rosala.....words have failed me.......!!!

x^3 = 2x+1    subtract everything on the right side to make it 0....so we have....

x^3 - 2x - 1  =  0     and from the Facror Theorem, we have that -1 is a root (solution)

Therefore, using a little synthetic division, we have

-1  [  1    0   - 2    -1 ]

            - 1     1     1

       -----------------------

       1     -1   -1     0

This tells us that the polynomial remaining after we divide  x^3 - 2x -1  by (x + 1)  = x^2 - x - 1

And setting this = 0, we have that x =

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.618\: \!033\: \!988\: \!749\: \!894\: \!8}}\\
{\mathtt{x}} = {\mathtt{1.618\: \!033\: \!988\: \!749\: \!894\: \!8}}\\
\end{array} \right\}$$

So these are the 3 solutions......BTW.....the last 2 solutions are better known as  "-phi" and "Phi"......

Mmmmm.....that's some interesting math, Anonymous....!!!

The answer is 0.3549995412350966

LOL!!!