Melody

OR maybe the asker may have meant

$$\\log_{256}x = 4\\\\
x=256^4$$

$${{\mathtt{256}}}^{{\mathtt{4}}} = {\mathtt{4\,294\,967\,296}}$$

Well Chris, what you have done looks all very impressive but I'm going to have a go at it myself.

I am treading on shaky ground here so if another mathematicians wants to correct me I shall not be too surprise.

let f(x)= $$\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{3}}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}$$, solve for (f^-1)'(3).

$$\\y=(0.25)x^3+x-1\\\\
$inverse function$\\\\
x=(0.25)y^3+y-1\\\\
\frac{dx}{dy^{-1}}=0.75y^2+1\\\\
\frac{dy^{-1}}{dx}=\frac{1}{0.75y^2+1}\\\\$$

----------------------------------------------

$$\\x=(0.25)y^3+y-1\\\\
When\;x=3\; \\\\
3=0.25*y^3+y-1\\\\
0=0.25y^3+y-4\\\\$$

Are you showing us what you have learned.  If so that is really good.

OR

Is there are question here?    

$$\\\frac{2}{x^{3/7}}+\frac{5}{x^{2/7}}-\frac{3}{x^{1/7}}=0\\\\
x^{3/7}*\left(\frac{2}{x^{3/7}}+\frac{5}{x^{2/7}}-\frac{3}{x^{1/7}}\right)=x^{3/7}*0\\\\
2+5x^{1/7}-3x^{2/7}=0\\\\
let\;y=x^{1/7}\\\\
2+5y-3y^2=0\\\\
3y^2-5y-1=0\\\\
y=\frac{5\pm\sqrt{25+12}}{6}\\\\
y=\frac{5\pm\sqrt{37}}{6}\\\\
x=y^{1/7}=\left(\frac{5\pm\sqrt{37}}{6}\right)^{1/7}$$

I haven't checked it but I think that is somewhere near correct. 

look here

there is a calc on the home page.

7/2 + 2109* (3218*32)

4*1.6*10^(9) = 6.4*10^9

Here are a couple more options.

$$\\ln(x^2)=1\\
2lnx=1\\
lnx=0.5\\
x=e^{0.5}\\$$