Alan

23^4 - 1 or

$${{\mathtt{23}}}^{{\mathtt{4}}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{279\,840}}$$

This answer is in base ten of course!

Warning!!

Beware of definite integration using trigonometric functions.  web2.0calc insists on treating the arguments to trig functions in degrees.  This leads to problems for definite integrals.  

For example 

$$\int_0^1{cos(x)}dx=0.841$$  

However, webcalc gives:

$$\int_{{\mathtt{0}}}^{{\mathtt{1}}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{x}}\right)}\right)}{d{\mathtt{x}}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{1}}^\circ\right)}$$

and 

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{1}}^\circ\right)} = {\mathtt{0.017\: \!452\: \!406\: \!437}}$$

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Indefinite integral:  int(f(x))   e.g.  int(x^2)

Definite integral: int(f(x),x=a..b)   e.g.  int(x^2,x=0..1)

Differentiation;  diff(f(x))   e.g.  diff(x^2)

(Thanks to Chris who just initiated my search for these!)

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e is the number

$${\mathtt{e}} = {\mathtt{2.718\: \!281\: \!828\: \!459\: \!045\: \!2}}$$ ... except that it goes on forever.

If you raise this number to the power of -1/24000 you get:

$${{\mathtt{e}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{24\,000}}}}\right)} = {\mathtt{0.999\: \!958\: \!334\: \!201\: \!376\: \!8}}$$ ... which also goes on forever!

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This is (nbr of ways of arranging 3 people in a group of 3 + nbr of ways of arranging 3 people in a group of 4 + nbr of ways of arranging 3 people in a group of 8)/(nbr of ways of arranging 3 people in a group of 15):

$${\frac{\left({\left({\frac{{\mathtt{3}}{!}}{({\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{4}}{!}}{({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{8}}{!}}{({\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}\right)}{{\left({\frac{{\mathtt{15}}{!}}{({\mathtt{15}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}}} = {\frac{{\mathtt{61}}}{{\mathtt{455}}}} = {\mathtt{0.134\: \!065\: \!934\: \!065\: \!934\: \!1}}$$

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Or do you mean  

$$2^{n-1}=8^n$$

in which case, because 8 = 2³

$$2^{n-1} = 2^{3n}$$

so n-1 = 3n 

-1 = 2n

n = -1/2

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No solution!

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(However, 8ⁿ - 1 = 2ⁿ does have a solution.)

1.  Write the circle equation in the form (x-xc)² + (y-yc)² = r²,  where (xc, yc) are the coordinates of the centre and r is the radius.  

2. Your circle equation can be written as (x-5)² + (y-11)² = 10², so the centre is at (5, 11).

3. Your straight line can be written in the form y1 = -(12/5)x - 54/5, so it has slope -12/5.

4. The slope of the straight line perpendicular to this is y2 = (5/12)x + k, where k is a constant.

5. For this second line to go through (5, 11) we must have 11 = (5/12)*5 + k so that k = 107/12

6. The equation of the second line is therefore y2 = (5/12)x + 107/12

7. This hits the first line when y1=y2.  The value of x at which this occurs is given by equating the two straight line equations:  -(12/5)x - 54/5 = (5/12)x + 107/12

8.  Rearrange as;  (5/12 + 12/5)x = -54/5 - 107/12  or (169/60)x = -1183/60 so that x = -1183/169 = -7

9. When x = -7 then y1 = y2 = -(5/12)*7 +107/12 = 72/12 = 6

10. The coordinates of P are therefore (-7, 6)

11. The distance between P and Q is given by √( (5 - (-7))² + (11 - 6)² ) = 25√2 ≈ 35.355

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Oops!  Ignore my result for 11.  It should be √( (5 - (-7))² + (11 - 6)² ) =√( (12²+5²) = √169 = 13. (thanks Chris!).

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