Question

log232 – 5log53 Help me please

Answer 1

Hello DeezyUs, first of all Welcome to OpenStudy! Thanks for coming to OpenStudy. I will guide you to the correct answer and yes, through the best possible way! But before I move on, I will like to know your ideas & work about this problem. Have you tried to solve it? What steps had you tried?

Answer 2

I am taking a online class and there is no teacher so I have no idea to solve this problem and there is no youtube video to help assist me in this problem.

Answer 3

Okay, I understand your problem! No worries, that's why OpenStudy has been invented - to help students with their problems.

Are you aware about Logarithm General Properties ?

Like : What is \(\log (\cfrac{a}{b})\) ?

Answer 4

No not at all. Please explain.

Answer 5

Oh! Well, no problem. I will try to help you out and make you understand about the general properties of Logarithm ... To start solving this problem, we must try to understand which properties we can apply here.

Answer 6

Following properties will be used while solving this question, please try to understand them so that we can make it through easily :

\(\boxed{\boxed{\color{blue}{\textbf{General Properties (to be used) of Logarithm} } } \\
\boxed{\bullet \sf \log (\cfrac{a}{b}) = \log a - \log b \\
\bullet \log(ab) = \log a + \log b \\
\bullet k \log a = \log (a^k)} \\
}\)

Answer 7

ok i understand that part but how do i solve the problem

Answer 8

Okay, we will touch only 1 term here, which is : \(\log(232)\) .

Can you do the prime factorization of 232 ?

Answer 9

Form "Prime Factorization" , I meant, can you tell me what are the prime factors of 232?

Answer 10

2 * 2 * 2 * 29

Answer 11

232 = 2 * 116

Answer 12

Great work. So, I can write \(\bf{\log{(\color{blue}{232})}} = \log{(\color{Blue}{2*2*2*29})}\)

(As I asked about "prime" factors, so let it be 2*2*2 * 29, it will make our work easier)

Answer 13

Now, I can write \(\bf{2*2*2}\) as \(\bf{2^3}\) , right?

Answer 14

yeah

Answer 15

So, we will right \(\bf \log{(\color{Blue}{2*2*2}*29)} \) as \(\bf \log(\color{blue}{2^3} *29)\)

Answer 16

ok

Answer 17

Now, can you tell me which property can be used to simplify the above term? (try to think 2^3 as "a' and 29 as "b")

Answer 18

14.5?

Answer 19

No, I meant to say that which property (check the box I made above) can be used to simplify that term?

Answer 20

Or theres not one

Answer 21

k log a

Answer 22

Yes, we will use that property but before that , we will use one more property

See, let us take 2^3 as a and 29 as b.

So, \(\log (2^3 \times 29) = \log(ab) \)

Answer 23

And since, \(\log(ab) = \log a + \log b\)

Are you getting my point?

Answer 24

A little

Answer 25

Okay, so, \(\log (2^3 \times 29) \) can be written as ?

(try to use log(ab) property)

Answer 26

log 2^3+ log 29

Answer 27

Excellent. Now, we will be using the property which you mentioned (k loga )

We will use it for the first term that is \(\log(2^3)\)

Answer 28

Can you simplify \(\log(2^3) \) using the property you mentioned?

Answer 29

ok so it would be log2^3+ 29

Answer 30

Yes it will be but we need to simplify \(\log(2^3)\) .

See, \(\log(a^k) = k\log (a)\)

Try to imagine 2 as a and 3 as k

Answer 31

??

Answer 32

I will try to give you an example :

\(\large{ \log(3^2) }\) can be written as \(\large{2 \log (3)}\)

Answer 33

What I did was just used the property :

\(\boxed{\large{\log(a^k) = k\log(a)}}\)

Can you try to use this in : \(\log (2^3)\) now?