Question

Solve for x, without a calculator

Answer 1

\[2\log _{6}(3x)=1\]

Answer 2

Start with the base change.

Answer 3

yep, I've done that.

Answer 4

\[\log _{6}(9x ^{2})=1\]
I've got up to that point.

Answer 5

Now convert into exponential form.

Answer 6

okay, this is gonna be fun, you answered fast

Answer 7

So I just divide both sides by 2

Answer 8

\[6=9x ^{2}\]

Answer 9

\[\rm \log _6 (9x^2) = 1 \implies 6^1 = 9x^2\]

Answer 10

Then I raise both sides 6=log6(3x)=6^(1/2)

Answer 11

3x=6^1/2

Answer 12

3x= sqrt 6

Answer 13

now divide both sides by 3

Answer 14

x=sqrt6/3 or x = sqrt(2/3)

Answer 15

Well, you should have let @JayDS do some work...

Answer 16

wait, I don't quite get your method @zordoloom, @ParthKohli how would I continue from 6=9x^2?

Answer 17

Find the square root of both sides. @JayDS

Answer 18

Or, you could divide both sides by 9. Then take the square root

Answer 19

A better way to start it was this:\[\log _6 (3x) = {1 \over 2}\]Now convert into exponential form.

Answer 20

actually, I got it now but you know how you square root 2/3?
it should be plus/minus square root 2/3 right? but I just take x for the positive value?

It's just that my cousin showed me a similar question and told me not to do that, so I was a bit confused.

Answer 21

Square root 2/3? You have this:\[\rm 6^{1 \over 2} = 3x\]

Answer 22

Now just divide both sides by 3 and you are done.

Answer 23

Or is that a different question?

Answer 24

yeh, I understand what you did from 6=9x^2
you squared 6/brought the squared over to the left which is the same as saying 6^1/2 but technically speaking it should be plus or minus 6?

Answer 25

x can't be negative, log of negative numbers is not defined

Answer 26

okok, got it, thanks.

Answer 27

Oh, that? Yup.

Answer 28

Better just write \(\rm \sqrt{2 \over 3 }\)

Answer 29

yep, I did.

Answer 30

Great! :)