Question

Expand each logarithm log3squareroot(2x) please help

Answer 1

\[\log_3 \sqrt{2x}\]
so it's
\[\large \log_3 (2x)^{\frac{1}{2}}\]
We know the logarithm property
\[\log_a b^c=c \times \log_a b\]
Using this here
\[\frac{1}{2}\large \log_3 2x\]
Do you get this?

Answer 2

not really idk how to get the anwser

Answer 3

Do you know logarithms and their properties ?

Answer 4

not well

Answer 5

Logarithms are defined as
\[\large \log_x y=z\]
This implies
\[\large y=x^z\]
Some important properties of logarithms
\[1)\ \log a^b= b\times \log a\]
\[2)\ \log ab=\log a+\log b\]
\[3)\ \log \frac ab=\log a-\log b\]
Base can be any number > 0 but not 1

Answer 6

well can you help me figure out the anwser?

Answer 7

@asdfqwer1234 I gotta go now, my friend will help you here
@apoorvk please help here

Answer 8

Hey asdfqwer1234 - did you follow what Ash said above?
If yes, good. If no, read it up very slowly, calmly and carefully. :)

Answer 9

The answer is actually staring at you :)

Answer 10

what i don't get it?

Answer 11

okay - \[(2x)^{\frac 1 2} = \sqrt{2x} = \sqrt2 \times \sqrt x\]

do you agree with this much?

Answer 12

yeah

Answer 13

so can you use the second property that Ash posted above - the one that says-
"log(a x b) = log(a) + log(b)"?

Here, your 'a' would be sqrt(2) and 'b' sqrt(x).

Answer 14

thanks

Answer 15

You sure you'll be able to do this stuff now?

Answer 16

no not really but i think i will figure it out

Answer 17

Well, just tag me in or anyone online (am going off right now) if you need more help!