How do I evaluate this log function?

Use the fact that

$$\log_b a = \frac{\log_c a}{\log_c b}$$

for any positive, real numbers a, b and c (with c > 1).

 

You use LaTeX, that is, the [ itex ][ /itex ] -tags (remove the spaces) . There's a guide in the forums somewhere if you're not familiar with it, I'll edit a link to it to this post if I find it. You can also open the "LaTeX Reference" by clicking the Σ in the toolbar.

If you don't have to write anything complicated, you can just use the quick symbols and x² and x₂ buttons in the toolbar. LaTeX looks a lot nicer & it's easier to read, though.

EDIT: Here's the LaTeX guide.

 

You could also approach this as follows:

Let [itex]\displaystyle y=(\log_{2}3)\,(\log_{3}4)\,(\log_{4}5)\,\dots\,( \log_{31}32) [/itex]

Then, [itex]\displaystyle 2^y=2^{(\,(\log_{2}3)\,(\log_{3}4)\,(\log_{4}5)\, \dots\,( \log_{31}32)\,)} [/itex]

By laws of exponents and the definition of a logarithm,

[itex]2^{(\,(\log_{2}3)\,(\log_{3}4)\,(\log_{4}5)\, \dots\,( \log_{31}32)\,)}[/itex]

[itex]=\left(2^{(\log_{2}3)}\right)^{(\,(\log_{3}4)\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}[/itex]

[itex]=3^{(\,(\log_{3}4)\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}[/itex]

[itex]=\left(3^{(\log_{3}4)}\right)^{(\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}[/itex]

[itex]=4^{(\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}[/itex]

etc.

[itex]=\left(31^{(\log_{31}32)}\right)[/itex]

[itex]=32[/itex]
​

 

So, just in case others misunderstand, [itex]2^y= 32= 2^5[/itex] and therefore, y= 5 as the original poster said.