Thread: A little bit lost with Log solving

1. 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

2. Re: A little bit lost with Log solving

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.

3. Re: A little bit lost with Log solving

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?

4. 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?

5. Re: A little bit lost with Log solving

Solve for a using the quadratic formula.

6. Re: A little bit lost with Log solving

Duh. I shoudl have seen that.

Cheers,

Perjac

7. Re: A little bit lost with Log solving

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 $a^2-3a=10$

8. Re: A little bit lost with Log solving

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

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

9. Re: A little bit lost with Log solving

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.

10. Re: A little bit lost with Log solving

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