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### Logarithm

```
Date: 4/6/96 at 22:41:35
From: Anonymous
Subject: Logarithm

How to solve this question?

LOG 2      156
```

```
Date: 7/31/96 at 15:7:21
From: Doctor Mike
Subject: Re: Logarithm

Hello,

I assume you mean "log base 2, of 156".  Let's just write log
for "log base 2", and a^b for a to the power b.  Since 2^7=128
and 2^8=256 you know that log(128)=7 and log(256)=8.  Because
156 is between 128 and 256, log(156) must be between 7 and 8.

My calculator does not do base 2 logs, but it does do natural
logs(base = e), so I'll use the relationship log(x)=ln(x)/ln(2).
So, log(156) = ln(156)/ln(2) which is 7.2854 approximately.

If you want to know why log(x) = ln(x)/ln(2) , here's a proof.
It must be shown that 2 to the power ln(x)/ln(2) gives just x.
I will be using the important fact that (a^b)^c = a^(b*c).

2^(ln(x)/ln(2)) = (e^ln(2))^(ln(x)/ln(2))
= e^(ln(2)*(ln(x)/ln(2)))
= e^ln(x)
= x

write back to us.  Best regards.

-Doctor Mike,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Logs

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