Log question

[BobbyJones]

I was getting stuck with this, trying to solve for x. The Log5 is base 5 as I couldnt do a little 5.

0.5Log5(x^2 - 1) = 0.25 + 0.5Log5(x-1)

I tried afew ways, like multipyling through by 2 to make

Log5(x^2 - 1) = 0.5 + Log5(x-1)

Log5(x^2 - 1) - Log5(x-1) = 0.5

Log5(x^2 - 1)
-------- = 0.5
(x-1)


(x^2 - 1)
-------- = 5^0.5
(x-1)


(x^2 - 1) = 5^0.5 (x-1)

x^2 - 1 = 2.23x- 2.23

x^2 = 2.23x- 1.23

x^2 - 2.23x + 1.23

divide by 2

0.5x^2 - 1.118x+ 0.615

From here I put it into the quadratic formula, but when I had two answers for x I tried them back in, and they didnt work out equal!

 

[galactus]

You have several errors throughout your attempt.

Perhaps try it this way by factoring \(\displaystyle x^{2}-1=(x+1)(x-1)\)

\(\displaystyle \frac{1}{2}log_{5}(x^{2}-1)=\frac{1}{4}+\frac{1}{2}log_{5}(x-1)\)

Factor the x^2-1 and use \(\displaystyle log(ab)=log(a)+log(b)\):

\(\displaystyle \frac{1}{2}log_{5}(x+1)+\frac{1}{2}log_{5}(x-1)=\frac{1}{4}+\frac{1}{2}log_{5}(x-1)\)

Note that the log(x-1) cancels and we are left with:

\(\displaystyle \frac{1}{2}log_{5}(x+1)=\frac{1}{4}\)

\(\displaystyle log_{5}(x+1)=\frac{1}{2}\)

\(\displaystyle x+1=\sqrt{5}\)

\(\displaystyle x=\sqrt{5}-1\)

 

[BobbyJones]

I didnt think of that, thanks.