help solving an equation involving logs

**absvalue** · October 6th 2009, 07:09 AM

$\sqrt{log(x) - 3} = log(x) - 3$

I'm not sure how to solve this. Logically, it would seem to me that $log(x) - 3$ must equal 0 or 1 for $\sqrt{log(x) - 3}$ to equal $log(x) - 3$ . I'm not sure how to prove it though. Could someone point me in the right direction?

**mathaddict** · October 6th 2009, 07:16 AM

Originally Posted by absvalue

$\sqrt{log(x) - 3} = log(x) - 3$

I'm not sure how to solve this. Logically, it would seem to me that $log(x) - 3$ must equal 0 or 1 for $\sqrt{log(x) - 3}$ to equal $log(x) - 3$ . I'm not sure how to prove it though. Could someone point me in the right direction?

HI

You can let log x = b

$\sqrt{b-3}=b-3$ .. square both sides and with some algebra

$b^2-7b+12=0$

$(b-4)(b-3)=0$

b=4 or b=3

Then log x =4 or log x =3

Solve for x .

**absvalue** · October 6th 2009, 07:27 AM

Thanks! That makes sense. So...

$\log(x) = 4 \Longleftrightarrow 10^4 = x$

or

$\log(x) = 3 \Longleftrightarrow 10^3 = x$

**mathaddict** · October 6th 2009, 07:39 AM

Originally Posted by absvalue

Thanks! That makes sense. So...

$\log(x) = 4 \Longleftrightarrow 10^4 = x$

or

$\log(x) = 3 \Longleftrightarrow 10^3 = x$

Absolutely

**earboth** · October 6th 2009, 07:45 AM

Originally Posted by absvalue

$\sqrt{log(x) - 3} = log(x) - 3$

I'm not sure how to solve this. Logically, it would seem to me that $log(x) - 3$ must equal 0 or 1 for $\sqrt{log(x) - 3}$ to equal $log(x) - 3$ . I'm not sure how to prove it though. Could someone point me in the right direction?

Here is another attempt:

If $z = \sqrt{\log(x)-3}$

then your equation becomes:

$z = z^2~\implies~z^2-z=0~\implies~z(z-1)=0~\implies~ z = 0~\vee~z=1$

Now re-substitute to calculate x.

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