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LynnM · Oct 5, 2010, 03:14 PM

Can someone just get me started on this question:

2(logx)^2 -3 -5logx = 0 (Solve for x.)

Not too sure how to begin this one. Thank you.

Unknown008 · Oct 6, 2010, 01:51 AM

You make use of a substitution to make it easier.

Let $y = log\ x$

$2y^2 -3 -5y = 0$

Rearrange:

$2y^2 -5y -3 = 0$

Solve this. After that, replace back y by log x.

Post what you get! :)

LynnM · Oct 6, 2010, 07:50 AM

Okay, here it is:

2y^2 - 5y - 3 = 0
(2y + 1) (y - 3) = 0

y = -1/2, 3

Here is where it gets tricky for me. Since a base can never be negative I'm not sure what to do about the -1/2. Would it still be a plausible solution since the entire logx may be a negative or is considered to be the base? This is the next step I took. Not too sure about it though.

2(-1/2)^2 - 3 - 5(-1/2) = 0
1/2 - 3 + 2.5 = 0
0 = 0
and
2(3)^2 - 3 - 5(3) = 0
18 - 3 - 15 = 0
0 = 0

Am I at least somewhat correct? (crossing my fingers!)

Unknown008 · Oct 6, 2010, 07:57 AM

Yes, but now, you must replace back y by log x, like this:

$y = -\frac12$

So,

$log\ x = -\frac12$

And:

$y = 3$

So,

$log\ x = 3$

Solve for x now :)

LynnM · Oct 6, 2010, 08:15 AM

Like this:

logx = -1/2
10^{-1/2} = x
x = 0.3162
and
logx = 3
10^3 = x
x = 1000

Then, plugged back into the equation:

2(log0.3162)^2 - 3 - 5log0.3162 = 0
0.5 - 3 + 2.5 = 0
0 = 0
and
2(log1000)^2 - 3 - 5log1000 = 0
18 - 3 - 15 = 0
0 = 0

So, because the 2 and the power 2 are placed outside the brackets surrounding the logx, they are applied after solving for log0.3162 and log1000, correct? And the because the 5 is not outside of any brackets it should be raised as a power then log1000^5 should be solved, correct?

Unknown008 · Oct 6, 2010, 08:21 AM

Right! :)

Well done!

LynnM · Oct 6, 2010, 08:24 AM

Yesss! -raises arms in triumphant relief-! Thanks a billion!

sandford11 · Oct 13, 2010, 12:26 AM

Good solutions there I like them.