1. ## Logs

Brand new to this so aplogies if anything is wrong:

Any help with this would be appreciated:

Solve x for the equation:

1/2 log5 (x^2-1) = 1/4 + 1/2log5(x-1)

The log5 is log to the base 5

Hope this makes sense

2. Originally Posted by wtopping
Brand new to this so aplogies if anything is wrong:

Any help with this would be appreciated:

Solve x for the equation:

1/2 log5 (x^2-1) = 1/4 + 1/2log5(x-1)

The log5 is log to the base 5

Hope this makes sense
Dear wtopping,

So your equation is, $\frac{1}{2}log_{5}(x^2-1)=\frac{1}{4}+\frac{1}{2}log_{5}(x-1)$

Am I correct??

3. Yes, exactly that, thanks for the quick reply,

Wayne

4. Originally Posted by wtopping
Yes, exactly that, thanks for the quick reply,

Wayne
Dear wtopping,

Let me give you some ideas....

First you can factorize $(x^2-1)$. Then try using the logarithm identity, $logA-logB=log\frac{A}{B}$

I hope you could continue from here!! In any case please don't hesitate to reply!!

5. Having come back to maths after a 15 year break i seem unable to see the woods for the trees. I think i can see that i need to bring the right hand side log to the left of = but stuck after that. I think the 1/2 is confusing me!!!

Its been a long time

6. Originally Posted by wtopping
Having come back to maths after a 15 year break i seem unable to see the woods for the trees. I think i can see that i need to bring the right hand side log to the left of = but stuck after that. I think the 1/2 is confusing me!!!

Its been a long time
Dear wtopping,

Ok. After taking the right hand side log to the left hand side try to apply the log identity I have given you. For this you have take 1/2 out of the bracket first.

7. Hello, wtopping!

Solve for $x\!:\;\;\tfrac{1}{2}\log_5(x^2-1) \:=\:\tfrac{1}{4}+ \tfrac{1}{2}\log_5(x-1)$

Note that: ${\color{blue}x \neq 1}$

Don't like fractions? . . . Eliminate them!

Multiply by 4: . $2\log_5(x^2-1) \;=\;1 + 2\log_5(x-1)$

. . $2\log_5(x^2-1) - 2\log_5(x-1) \;=\;1 \quad\Longrightarrow\quad 2\bigg[\log_5(x^2-1) - \log_5(x-1)\bigg] \;=\;1$

. . $2\log_5\left[\frac{x^2-1}{x-1}\right] \;=\;1 \quad\Longrightarrow\quad 2\log_5\left[\frac{(x-1)(x+1)}{x-1}\right] \;=\;1$

. . $2\log_5(x+1) \;=\;1 \quad\Longrightarrow\quad \log_5(x+1) \:=\:\tfrac{1}{2}$

. . $x + 1 \;=\;5^{\frac{1}{2}} \quad\Longrightarrow\quad x \;=\;\sqrt{5} - 1$