# help solving an equation involving logs

• Oct 6th 2009, 07:09 AM
absvalue
help solving an equation involving logs
$\displaystyle \sqrt{log(x) - 3} = log(x) - 3$

I'm not sure how to solve this. Logically, it would seem to me that $\displaystyle log(x) - 3$ must equal 0 or 1 for $\displaystyle \sqrt{log(x) - 3}$ to equal $\displaystyle log(x) - 3$. I'm not sure how to prove it though. Could someone point me in the right direction?
• Oct 6th 2009, 07:16 AM
Quote:

Originally Posted by absvalue
$\displaystyle \sqrt{log(x) - 3} = log(x) - 3$

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

HI

You can let log x = b

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

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

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

b=4 or b=3

Then log x =4 or log x =3

Solve for x .
• Oct 6th 2009, 07:27 AM
absvalue
Thanks! That makes sense. So...

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

or

$\displaystyle \log(x) = 3 \Longleftrightarrow 10^3 = x$
• Oct 6th 2009, 07:39 AM
Quote:

Originally Posted by absvalue
Thanks! That makes sense. So...

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

or

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

Absolutely (Clapping)
• Oct 6th 2009, 07:45 AM
earboth
Quote:

Originally Posted by absvalue
$\displaystyle \sqrt{log(x) - 3} = log(x) - 3$

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

Here is another attempt:

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

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