Is My Solution to This Log Question Correct?

[elctronoob]

Hi folks,

I'm revisiting logs for the first time in a long time through distance education and I was wondering if someone could have a look over a question I've answered and let me know if I've done it correctly or if I'm way off please

Find x if Log3(10x – 1) – 2 = 2log3x

I instantly divide by common log of log3 and work out from there

(10x-1) - 2 / 2 = x
(5x- 1/2) -1 = x
5x - 1/2 = x + 1
5x - x = 1 + 1/2
4x = 3/2
x = 3/8

Any help would be greatly appreciated

 

[shadowshed]

By 2Log3, does it mean that it's a log with basis of 2 or 2*Log3?

 

[elctronoob]

its written as 2log₃x, so I take it to mean 2 * log₃X

 

[Samy_A]

If it means
Find $x$ if $\log_3(10x-1) -2=2*\log_3x$,
I don't understand what you did.

You could take all the $log_3$ terms together, making use of:
$\log_3(a*b)=\log_3a+\log_3b$
$\log_3(a/b)=\log_3a-\log_3b$
$2*\log_3a=\log_3a^2$
$3^{log_3a}=a$

and then solve for x.

 

[shadowshed]

Since 2log3x = 2 (Log3 + Log x)
which means there will 2 Log 3, so there would be still one Log 3 remaining.

Or you can just actually change the 2Log3x to Log(3x)^2, and then manipulate the left side of equation to be Log, and equals them.

 

[elctronoob]

Thanks everyone for all the help...its been far too long since I've looked at any of this sort of stuff...
I now have it at
(log₃10x - log₃1) - 2 = log₃x²
Then divide out the log₃
10x - 1 - 3 = x²
0 = x² -10x + 3

I've obviously done something wrong again as this gives pretty silly results :(

 

[shadowshed]

You've got to make (log310x - log31) - 2 into one Log3. You don't really divide out the log, but comparing the values inside the logs. They have to be equals, since the logs would result the same.

 

[Samy_A]

Thanks everyone for all the help...its been far too long since I've looked at any of this sort of stuff...
I now have it at
(log₃10x - log₃1) - 2 = log₃x²

How did you get the expression to the left?
Note that $\log_3(10x-1)\neq\log_3(10x)-\log_3(1)$

Then divide out the log₃

Not sure what you mean by "divide out the log₃"

 

[jedishrfu]

You start with finding what the value of 2 is in $log_3$

$log_3(3)=1$ and $log_3(9)=log_3(3^2)=2$

So the RHS becomes $log_3(10x - 1) - log_3(9) = log_3( (10x - 1) * 1/9)$

and go from there.

 

[elctronoob]

How did you get the expression to the left?
Note that $\log_3(10x-1)\neq\log_3(10x)-\log_3(1)$

Not sure what you mean by "divide out the log₃"

i'm just going round in circles here and getting nowhere I'm afraid
$\log_3(10x-1)=((log_3(10) + log_3(x)) / log_3(1))$

I understood when i had a common log base, i could divide all by that log base
hence ((10 + x) / 1) - 2 = x²
0 = x² - x - 8

which doesn't give reasonable answers so i must have messed up somewhere again

 

[jedishrfu]

You're not dividing by log, you have to get that understanding straight first to solve this problem:

$log3(10x−1) =/= ((log3(10)+log3(x))/log3(1))$

 

[jedishrfu]

Khan's academy videos may help you get over your misconceptions:

https://www.khanacademy.org/math/al...tions/introduction-to-logarithms/v/logarithms

 

[elctronoob]

Khan's academy videos may help you get over your misconceptions:

https://www.khanacademy.org/math/al...tions/introduction-to-logarithms/v/logarithms

Thanks jedishrfu and everyone else...I'm going to walk away from it for a while n watch that video... n come back to it later tonight... 15 year old version of me would be lolling his head off at me now...if lolling was a thing back then

 

[elctronoob]

i'm just going round in circles here and getting nowhere I'm afraid
$\log_3(10x-1)=((log_3(10) + log_3(x)) / log_3(1))$

I understood when i had a common log base, i could divide all by that log base
hence ((10 + x) / 1) - 2 = x²
0 = x² - x - 8

which doesn't give reasonable answers so i must have messed up somewhere again

Right folks, I've gone away n come back with an improved effort i think...

log₃(10x-1) - 2 = log₃(x²)

=log₃((10x-1)/x²) = 2

= (10x-1)/x² = 3² = 9

9x² -10x + 1 = 0

=>x = 1, x = 1/9

Does this look correct to you guys or am I missing something again?

 

[Samy_A]

Right folks, I've gone away n come back with an improved effort i think...

log₃(10x-1) - 2 = log₃(x²)

=log₃((10x-1)/x²) = 2

= (10x-1)/x² = 3² = 9

9x² -10x + 1 = 0

=>x = 1, x = 1/9

Does this look correct to you guys or am I missing something again?

Looks correct.

 

[elctronoob]

Looks correct.

Thanks very much for your help Samy and the rest of you fine people, much appreciated