# A little bit lost with Log solving

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• July 31st 2011, 02:41 PM
perjac
A little bit lost with Log solving
Hi All,

Bit lost here, tried all sorts of converting but can't quite work it out, just need a push in the right direction.

(lnx)^2-ln(x^3)=10

the square outside the brackets is confusing me.\

Cheers,

Perjac
• July 31st 2011, 02:47 PM
Archie Meade
Re: A little bit lost with Log solving
Quote:

Originally Posted by perjac
Hi All,

Bit lost here, tried all sorts of converting but can't quite work it out, just need a push in the right direction.

(lnx)^2-ln(x^3)=10

the square outside the brackets is confusing me.\

Cheers,

Perjac

Use

$ln\left(a^b\right)=blna$

for the middle term
and have a look at what kind of resultant equation you have.
• July 31st 2011, 03:12 PM
Plato
Re: A little bit lost with Log solving
Quote:

Originally Posted by perjac
$(\ln(x))^2-\ln(x^3)=10$

You should know that $\ln(x^3)=3\ln(x).$
Let $z=\ln(x)$ and you get $z^2-3z-10=0$ to solve.

Can you go forward?
• July 31st 2011, 03:30 PM
perjac
Re: A little bit lost with Log solving
I guess this then becomes an algebra problem (let lnx = a)

a^2 - 3a = 10

though I am not sure how to solve for a. Also is there a link to how to use teh math tags?
• July 31st 2011, 03:50 PM
pickslides
Re: A little bit lost with Log solving
Solve for a using the quadratic formula.
• July 31st 2011, 03:52 PM
perjac
Re: A little bit lost with Log solving
Duh. I shoudl have seen that.

Cheers,

Perjac
• July 31st 2011, 03:53 PM
Plato
Re: A little bit lost with Log solving
Quote:

Originally Posted by perjac
I guess this then becomes an algebra problem (let lnx = a) a^2 - 3a = 10
though I am not sure how to solve for a. Also is there a link to how to use teh math tags?

I do not believe you there.
If you cannot solve that simple quadratic equation, you have no business trying to solve this question.

As for the use of LaTeX:
Why not learn to post in symbols? You can use LaTeX tags
Example
$$a^2-3a=10$$ gives $a^2-3a=10$
• July 31st 2011, 05:02 PM
Archie Meade
Re: A little bit lost with Log solving
Quote:

Originally Posted by perjac
I guess this then becomes an algebra problem (let lnx = a)

a^2 - 3a = 10

though I am not sure how to solve for a. Also is there a link to how to use teh math tags?

You can just find the factors of 10 that differ by 3,
which are 5 and 2.

$a^2-3a-10=(a-5)(a+2)$

and if this is zero, then either factor may be zero.

Or

$a^2-3a=a(a-3)=10=5(2) \;\;or\;\; (-2)(-5)$
• July 31st 2011, 05:22 PM
perjac
Re: A little bit lost with Log solving
Quote:

Originally Posted by Plato
I do not believe you there.
If you cannot solve that simple quadratic equation, you have no business trying to solve this question.

As for the use of LaTeX:
Why not learn to post in symbols? You can use LaTeX tags
Example
$$a^2-3a=10$$ gives $a^2-3a=10$

yeah brain fart there. My bad. Typed before thinking.

Thanks for the help guys.
• August 6th 2011, 08:17 PM
nkenji
Re: A little bit lost with Log solving
Quote:

Originally Posted by Archie Meade
You can just find the factors of 10 that differ by 3,
which are 5 and 2.

$a^2-3a-10=(a-5)(a+2)$

and if this is zero, then either factor may be zero.

Or

$a^2-3a=a(a-3)=10=5(2) \;\;or\;\; (-2)(-5)$

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