Exact Value of 6561(1/ln(9)) Using Logarithmic Form

  • Thread starter Thread starter imull
  • Start date Start date
imull
Messages
40
Reaction score
0

Homework Statement


Find the exact value of the expression: 6561(1/ln(9))

Homework Equations


logax=(1/ln(a))(ln(x))

The Attempt at a Solution


I thought that I would start it by putting the expression into logarithmic form: log6561x=1/(ln(9)). I am not sure where to go from here though.
 
Physics news on Phys.org
Try a different base.
 
Such as?
 
imull said:
Such as?

You already have an exact value. It's just what's printed there. You can express it in other ways that don't look terribly much simpler. I suspect that the log in the exponent might be intended to be a different base. Is that the exact problem? Try factoring 6561 to see why I might think so.
 
Last edited:
Dick said:
You already have an exact value. It's just what's printed there.
Exactly.

You can express it in other ways that don't look terribly much simpler.
It can be expressed in a much more compact form than 6561(1/ln 9).

Try factoring 6561 to see why I might think so.
That's a very good suggestion.
 
Thank you both very much. I'll work on it.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top