Logarithm Easy Question Help!

**r2d2** · November 22nd 2009, 06:39 AM

If $lnx^2= 3x$ , Find the Value of X.

I am missing a major property. Could someone help me 0ut on this one?!

Thank you

**e^(i*pi)** · November 22nd 2009, 06:42 AM

Originally Posted by r2d2

If $lnx^2= 3x$ , Find the Value of X.

I am missing a major property. Could someone help me 0ut on this one?!

Thank you

Do you mean $ln(x^2)$ or $[ln(x)]^2$

Either way you will need to use an iterative solution

**r2d2** · November 22nd 2009, 06:43 AM

I mean $ln(x^2)$

**e^(i*pi)** · November 22nd 2009, 06:48 AM

Originally Posted by r2d2

I mean $ln(x^2)$

There are actually no real (and possibly no complex) solutions for this equation because the two lines do not meet - see the attached graph

**r2d2** · November 22nd 2009, 06:49 AM

well I have the answer is equal to $e^3/2$ . Or e^(3/2)

**e^(i*pi)** · November 22nd 2009, 06:54 AM

Originally Posted by r2d2

well I have the answer is equal to $e^3/2$ . Or e^(3/2)

If you sub that back into the original equation you do not get 0

$<br /> 2ln(e^{1.5}) \neq 3e^{1.5}<br />$

**r2d2** · November 22nd 2009, 06:59 AM

Ok I must have copied the problem wrong.

Here it is: not simplified.

$-x-2x+2xlnx=0$

Solve for x.

**skeeter** · November 22nd 2009, 07:09 AM

Originally Posted by r2d2

Ok I must have copied the problem wrong.

Here it is: not simplified.

$-x-2x+2xlnx=0$

Solve for x.

$-x - 2x + 2x\ln{x} = 0$

$-3x + 2x\ln{x} = 0$

$x(-3 + 2\ln{x}) = 0$

$x = 0$ is an invalid solution because of domain reasons.

$\ln{x} = \frac{3}{2}$

$x = e^{\frac{3}{2}}$

**iPod** · November 22nd 2009, 07:17 AM

i also got e^(3/2), you just copied it down wrong the first time,
so ur correct
as;
-x-2x+2xlnx=0
-3x+2xlnx=0
2xlnx=3x
lnx=3x/2x
lnx=3/2
x=e^(3/2)

**r2d2** · November 22nd 2009, 07:21 AM

Great! Thank you everybody

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