Calculating base 2 log in TI-83 Plus?

Question

Any way to do it?

Answer 1

Let log_2(x) denote the log of x with base 2.

Use the identity that log_2(x) = ln(x) / ln(2). This identity holds because:

Recall that, by definition, 2^(log_2(x)) = x.

ln(x) = ln(2^(log_2(x))) = (log_2(x)) (ln(2)), so dividing through by ln(2) shows the result.

Answer 2

For the best answers, search on this site https://shorturl.im/awAPG

The only way you can do this is by simplifying that logarithm into this form: log(2)/log(1.041) Plug that into the TI-83 Calculator and the answer should come up to be 17.2503. It's actually log(2)/log(1.041), not the other way around as the other person put it. Note that the TI-83 Plus Calculator cannot calculate the logarithm of 2 with base 1.041 without first simplifying the logarithm. The only calculators made by Texas Instruments capable of performing this operation exactly as shown are the TI-89 and more advanced calculators by Texas Instruments such as the Voyage 200.

Answer 3

This Site Might Help You.

RE:
Calculating base 2 log in TI-83 Plus?
Any way to do it?

Answer 4

You can actually use a normal calculator for that just use change of base log [base a] b = log b / log a So log[base 1.041](2) = log 2 / log 1.041 Now you can use your calculator You should get 17.25 Sorry i didn't answer your question. I really don't know what a TI-83 calculator is .

Answer 5

Antilog On Ti-83

Answer 6

lon_2(x) = log(x)/log(2). e.g. 2^6 = 64
log_2(64) = log(64)/log(2) =
1.8129133566428555739927662632178/
0.30102999566398119521373889472449 = 6 using base 10 (common) logs. Also ln(64)/ln(2) =
4.1588830833596718565033927287491/
0.69314718055994530941723212145818 = 6 using base e (natural) logs. In general, the log to any base b of some number x can be found as log_b(x) = log(x)/log(b) or ln(x)/ln(b).


Doug

Answer 7

I'm not sure. I heard TI-83 only does base 10.

Answer 8

log number / log 2 will give you the logarithm in base 2.

You can easily program the above in your calculator.

I do not understand why people want to take such a simple problem and make it so darn complex - maybe if I wasn't 3 times the age of these posters, I could understand. Simplicity is the recommended path....

Answer 9

You cant program it do another base, but it is possible.

If you have
Log (base a) of X

Type in:
log x / log a

Then it will work. :) For any base.