how to check that n is a power of 3 [newbie]

[Candide Voltaire]

I need to check whether a number is a power of 3.
I tried the following:
isinteger(log(n)/log(3));

this doesn't work however (e.g. with n=27), probably because of the
floating point representation used internally in octave

Can anyone here suggest how to check whether n is a power of 3?

regards,
Candide

[Christoph Brinkhaus]

Candide Voltaire <***@gmail.com> schrieb:

Dear Candide,

Post by Candide Voltaire
I need to check whether a number is a power of 3.
isinteger(log(n)/log(3));

I am not 100% sure about the math expression.
For example 8 should be 2 in the power of 3.
Here log(8)/log(2) = 3.

Post by Candide Voltaire
this doesn't work however (e.g. with n=27), probably because of the
floating point representation used internally in octave

How about checking the expression

log(n)/log(3) - floor( log(n)/log(3) ).

The expression is 0.0 if it meets
your intention. It is > 0.0 otherwise.

Post by Candide Voltaire
Can anyone here suggest how to check whether n is a power of 3?

I am not sure if I have fully understood your question.
At least I hope I could provide some useful suggestion.

Post by Candide Voltaire
regards,
Candide

Kind regards,

Christoph

[Candide Voltaire]

Post by Christoph Brinkhaus
Dear Candide,

Post by Candide Voltaire
I need to check whether a number is a power of 3.
isinteger(log(n)/log(3));

I am not 100% sure about the math expression.
For example 8 should be 2 in the power of 3.
Here log(8)/log(2) = 3.

Post by Candide Voltaire
this doesn't work however (e.g. with n=27), probably because of the
floating point representation used internally in octave

How about checking the expression
log(n)/log(3) - floor( log(n)/log(3) ).
The expression is 0.0 if it meets
your intention. It is > 0.0 otherwise.

Post by Candide Voltaire
Can anyone here suggest how to check whether n is a power of 3?

I am not sure if I have fully understood your question.
At least I hope I could provide some useful suggestion.

Post by Candide Voltaire
regards,
Candide

Kind regards,
Christoph

thanks this is a useful answer

regards,
candide