Simplifying a Logarithm

[Relz]

Find the value of the following: 2log₃9
Leave answer as a fraction where appropriate and do not change the base.

does it then simplify to log₃ 9² = log₃ 81 ?

 

[soroban]

Hello, Relz!

Find the value of: .\(\displaystyle 2\log_39\)
Leave answer as a fraction where appropriate and do not change the base.

Does it then simplify to: .\(\displaystyle \log_3(9^2) \,=\, \log_3(81)\, ?\) . Yes!

Can you evaluate that?

 

[Relz]

Hi! Im not sure what you mean by evaluate, do you mean solve it fully? Just another quick question: why isn't it 2 log₃ 3² = 2 (1) (2)= 4 ?

 

[pka]

Hi! Im not sure what you mean by evaluate, do you mean solve it fully? Just another quick question: why isn't it 2 log₃ 3² = 2 (1) (2)= 4 ?

Surely you know that \(\displaystyle 9^2=3^4\).

So \(\displaystyle \log_3(9^2)=\log_3(3^4)=4.\)

 

[mmm4444bot]

Im not sure what you mean by evaluate

"Evaluate" is a verb. It means "to find the value of".

The exercise asks you to find the value of the expression 2 log₃(9).

Hence, you need to evaluate the expression log₃(81).

(Knowing the multiplication table allows one to do this evaluation mentally.)

, do you mean solve it fully?

FYI: In math, the verb "solve" generally applies to equations, not expressions.

why isn't it 2 log₃3² = 2 (1) (2)

How did you determine that log₃(9) is the same as (1)(2) ?

I mean, if you can tell me why you think it is, then I'll be in a position to tell you why it isn't. :cool:

 

[Relz]

How did you determine that log₃(9) is the same as (1)(2) ?

I mean, if you can tell me why you think it is, then I'll be in a position to tell you why it isn't. :cool:

I think log₃ 9 = log₃ (3²) = 2
Also, log₃ (3)= 1

I guess a better way to put it is log₃ (9) is the same as 2.

 

[mmm4444bot]

Yeah, the end value is correct, but I'm trying to understand why you wrote 2 as (1)(2).

Maybe I do not understand what you're trying to ask, in your question "why isn't it … 2(1)(2)?"

In other words, I'm thinking that your pronoun "it" refers more to process versus end results.

 

[Relz]

Yeah, the end value is correct, but I'm trying to understand why you wrote 2 as (1)(2).

Maybe I do not understand what you're trying to ask, in your question "why isn't it … 2(1)(2)?"

In other words, I'm thinking that your pronoun "it" refers more to process versus end results.

Sorry, I do understand the 2 as (1)(2) is confusing, that was an error on my end! I would like to know which way I am to do it, whether it's the first way where I end up with the 9² or the second way where I end up with 4 as my final answer.

 

[mmm4444bot]

In this thread, I'm not sure what is the "first way" and what is the "second way". :-(

After we arrive at the simplification log₃(9²), then it's simply a matter of asking our self the question: "3 raised to what exponent equals 81?"

The multiplication table leads us to realize that this exponent must be 4.

Alternatively, you could simplify differently at the start.

2 log₃(9) = 2 log₃(3²) = 2(2) = 4

 

[Relz]

In this thread, I'm not sure what is the "first way" and what is the "second way". :-(

After we arrive at the simplification log₃(9²), then it's simply a matter of asking our self the question: "3 raised to what exponent equals 81?"

The multiplication table leads us to realize that this exponent must be 4.

Alternatively, you could simplify differently at the start.

2 log₃(9) = 2 log₃(3²) = 2(2) = 4

OH! Okay, so actually, there are two ways to arrive at the final answer of 4?
Also, sorry about even more confusion, I have been at this assignment for a few hours and I definitely need to take a break...my mind is getting foggy.

 

[mmm4444bot]

so actually, there are two ways to arrive at the final answer of 4?

There are more than two ways!

 

[lookagain]

... why isn't it 2 log₃ 3² = 2 (1) (2)= 4 ?

This is my take on this. However, if I am stating something that does not follow
and/or was not on the OP's mind, then void the following:


\(\displaystyle 2\log_3{3^2} =\)


\(\displaystyle 2(2)\log_3{3} \ \ or\)


\(\displaystyle 2(\log_3{3})2 \ =\)


\(\displaystyle 2(1)(2) \ =\)


\(\displaystyle 4\)

 

[Relz]

This is my take on this. However, if I am stating something that does not follow
and/or was not on the OP's mind, then void the following:


\(\displaystyle 2\log_3{3^2} =\)


\(\displaystyle 2(2)\log_3{3} \ \ or\)


\(\displaystyle 2(\log_3{3})2 \ =\)


\(\displaystyle 2(1)(2) \ =\)


\(\displaystyle 4\)

That's pretty much what I was thinking...but written out! Thanks

 

[mmm4444bot]

That's pretty much what I was thinking...but written out!

Egads!

I wish you had written out your reasoning, when you asked the question why it could not be done this way.

Your reasoning was valid; I must eat some of my prior words (in the corner).