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
Hello, wtopping!
Solve for $\displaystyle x\!:\;\;\tfrac{1}{2}\log_5(x^2-1) \:=\:\tfrac{1}{4}+ \tfrac{1}{2}\log_5(x-1)$
Note that: $\displaystyle {\color{blue}x \neq 1}$
Don't like fractions? . . . Eliminate them!
Multiply by 4: .$\displaystyle 2\log_5(x^2-1) \;=\;1 + 2\log_5(x-1) $
. . $\displaystyle 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 $
. . $\displaystyle 2\log_5\left[\frac{x^2-1}{x-1}\right] \;=\;1 \quad\Longrightarrow\quad 2\log_5\left[\frac{(x-1)(x+1)}{x-1}\right] \;=\;1$
. . $\displaystyle 2\log_5(x+1) \;=\;1 \quad\Longrightarrow\quad \log_5(x+1) \:=\:\tfrac{1}{2} $
. . $\displaystyle x + 1 \;=\;5^{\frac{1}{2}} \quad\Longrightarrow\quad x \;=\;\sqrt{5} - 1$