Question

how do you solve 2 log 3-log x = 2?

Answer 1

have you covered logarithms yet?

Answer 2

you know the property of logarithms ?

Answer 3

no not really like we have just started

Answer 4

2log 3 =log 3^2
yes ?

Answer 5

First, see that 2log(3) is equal to log(3^2). Therefore we have log(3^2) - log (x) (we know that 3^2 is 9. Then, we combine the logs making log (9/x). Finally we must turn 2 into a log. Since we know that log (10) = 1, then log (10^2) = 2. so we have Log (9/x)=Log(10^2). since both are in log form, we cancel the logs and get 9/x=10^2.

Answer 6

oh i still dont understand tbqh

Answer 7

and log a - log b= log a/b

Answer 8

To answer these questions I really suggest that you go over logarithm rules:
log (a) - log (b)= log(a/b)
log (a) + log (b)= log(a*b)
log (a) = log (b) is the same as a=b

Answer 9

hmmm so.... how are expected to do this exercise without having covered them yet?

Answer 10

the simple log
do you know what base has ?

Answer 11

Also, if the problem writes log (a), it means \[\log_{10}a \]

Answer 12

ok my teacher in school just started log and gave us some problems for homework and i wasnt too sure of how to go about it, my teahcer gave me a textbook and basically said learn it yourself

Answer 13

no i dont know what simple log is

Answer 14

depend how you have learned because when i have learned about so then lg was the base 10 , and log was the base 2

Answer 15

I imagine that the teacher hoped that you taught these rules yourself. To be honest, they are very easy to remember (thankfully).

A simple log is written simply as "log a" but really means \[\log_{10}(a) \]

Answer 16

yeah, idk anyways thanks to all for your help, im sure ill find another example in the book soon thanks hope yall have a great day and thanks again

Answer 17

so there will be

log 9/x = 2

now you need to know what is the base because you need rewriting the 2 using this logarithm

Answer 18

\[\log_{10} (a) \] simply means 10^x = a

Answer 19

ok

so than you can writing the 2 like log 10^2

yes ?