How do I evaluate this log function?

[feihong47]

Homework Statement

Homework Equations

The Attempt at a Solution

I assume I can't use a calculator obviously.. so I'm quite stuck. The answer is 5, but I have no idea how to get that.

(log₂3)(log₃4)(log₄5) ... (log₃₁32)

 

[Mute]

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).

 

[feihong47]

That was a fast response, and a fact that I didn't know. How do I create equations on this forum BTW?

 

[DeIdeal]

That was a fast response, and a fact that I didn't know. How do I create equations on this forum BTW?

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.

 

[SammyS]

Homework Statement

Homework Equations

The Attempt at a Solution

I assume I can't use a calculator obviously.. so I'm quite stuck. The answer is 5, but I have no idea how to get that.

(log₂3)(log₃4)(log₄5) ... (log₃₁32)

You could also approach this as follows:

Let $\displaystyle y=(\log_{2}3)\,(\log_{3}4)\,(\log_{4}5)\,\dots\,( \log_{31}32)$

Then, $\displaystyle 2^y=2^{(\,(\log_{2}3)\,(\log_{3}4)\,(\log_{4}5)\, \dots\,( \log_{31}32)\,)}$

By laws of exponents and the definition of a logarithm,

$2^{(\,(\log_{2}3)\,(\log_{3}4)\,(\log_{4}5)\, \dots\,( \log_{31}32)\,)}$

$=\left(2^{(\log_{2}3)}\right)^{(\,(\log_{3}4)\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}$

$=3^{(\,(\log_{3}4)\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}$

$=\left(3^{(\log_{3}4)}\right)^{(\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}$

$=4^{(\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}$

etc.

$=\left(31^{(\log_{31}32)}\right)$

$=32$
​

 

[HallsofIvy]

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

 

[feihong47]

Very nice approach. Thanks!