Log with base 3 or other bases

[JakePearson]

hey guys, i have a question, then i will show my solution, is there any chance you guys could help me find an easier way to get to the answer, cheers

question:
log_3 3x

answer:
log_a (yx) = log_a (y) + log_a (x)
log_3 (3) + log_3 (x)
log_a (y) = x -> y = a^x -> log_a (a) = 1 -> y = a^x -> x = 1

1 + log_3 X

is there an easier way of getting to this answer

 

[rock.freak667]

That is the way to get the answer. You don't need to put the second to last line though in most cases unless you need to write down the rules when you apply them.

 

[tomcue]

?

Hi there! It looks like you're trying to simplify the expression log_3 3x. You're definitely on the right track by using the property log_a (yx) = log_a (y) + log_a (x) to break down the expression. However, there is a simpler way to get to the answer. Since we know that log_3 3 = 1 (since 3^1 = 3), we can simplify the expression to just 1 + log_3 x. This means that the final answer is just 1 + log_3 x. I hope this helps! Let me know if you have any other questions.