Question

Find the value of x in each of the following equations. @Kainui

Answer 1

\[3^x-6^{3x+1}=0\]

Answer 2

\[3^x=6^{3x+1}\]

Answer 3

I suggest taking log base 3 to both sides. =)

Answer 4

\[\log _{3}3^x=\log _{3}6^{3x+1}\]

Answer 5

What is log base 3 of 9?

Answer 6

\[\log _{3}3^2\]

Answer 7

2

Answer 8

@Kainui

Answer 9

Right, so repeat what you've done here except replace "2" with "x" since that is the only difference between 9 and 3^x.

Answer 10

I would take the ln or log10 of both sides.

Answer 11

@phi the only reason I am suggesting log base 3 is because it would make the answer fit with what he already know from the previous question he asked: http://openstudy.com/users/marc_#/updates/54df4353e4b07f4a24bf9d4c

Answer 12

\[2=\log _{3}6^{3x+1}\]

Answer 13

\[3^2=6^{3x+1}\]

Answer 14

\[9=6^{3x}\times6^1\]

Answer 15

No, that's not quite what I'm saying, I'm not saying x=2. What you have written is wrong. I'm saying the method you used to get 2 will be the same: So look at the similarities her. \[\Large \log_3 3^2 = 2 \log_3 3 =2 \\ \Large \log_33^x=x \log_3 3 = x\]

Answer 16

\[x=\log _{3}6^{3x+1}\]

Answer 17

yes. can you do the same thing to the right hand side?

Answer 18

\[x=3x+1\log _{3}6\]\[x=3x+1\log _{3}(3\times2)\]

Answer 19

yes, but put parens around 3x+1 because (3x+1) should be "all together"

Answer 20

u mean like this? \[x=3x+1\log _{3}3\times3x+1\log _{3}2\]

Answer 21

like this
\[ x=(3x+1)\log _{3}(3\times2) \]

Answer 22

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

Answer 23

oops we have to be careful
\[ x=(3x+1)\log _{3}(3\times2) \\
x=(3x+1)( \log _{3}(3) + \log _{3}(2)) ) \]
the log(3) is 1 so this is
\[ x=(3x+1)( 1+ \log _{3}(2)) ) \]
now "distribute" the (3x+1)

Answer 24

you should get
\[ x=(3x+1)+ (3x+1)\log _{3}(2) \]

Answer 25

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

Answer 26

no. we can't add 3x+3x (because of the log(2) )
we can "distribute the log(2)"

\[ x = 3x+1 + \log_3(2) \cdot 3 x + \log_3(2) \]
I would subtract x from both sides:
\[ x-x = 3x-x +1 + \log_3(2) \cdot 3 x + \log_3(2) \]

Answer 27

\[0=2x+1+\log _{3}2\times3x+\log _{3}2\]

Answer 28

ok, and if we "change the order of multiply" and how we add, we can write it as:
\[ 0 = 2x + 3 \log_3(2) x + 1 + \log_3(2) \]

Answer 29

we are solving for x, so move the terms without x to the other side
i.e. add -1 - log(2) to both sides

Answer 30

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

Answer 31

next, factor x out of each term on the right side

Answer 32

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

Answer 33

now divide both sides by (2+3log(2) )

that gives you x

Answer 34

you have the value of log_3(2) from your previous problem
or use
\[ \log_3 2 = \frac{\ln 2}{\ln 3} \]

Answer 35

you should get
\[ x = \frac{-1 - \log_3 2}{2+3 \log_3 2} \]
and \( \log_3 2 \approx 0.63093 \)

Answer 36

can you find x?

Answer 37

x=-0.419

Answer 38

Thnx @phi and @Kainui

Answer 39

yes, that looks good.