$\displaystyle log_5(x+7) - log_(\sqrt5) (x+1)=1$
Er I think my question wasn't very clear because of LaTex...
The 1/2 is actually the power of the second log5 which means it is $\displaystyle log_5(x+7) - log_(\sqrt5) (x+1)=1$
The log having a base of square root is that making it difficult for me
Consider first $\displaystyle log_4(16)=2$, however if you want to take the square root of the base and preserve the same answer, you would have to take the square root of 16 as well, ergo $\displaystyle log_2(\sqrt16)=2$
Therefore similarly you can rewrite the LHS of the original equation as:
LHS=$\displaystyle log_5(x+7)-log_5((x+1)^2)$
hence solve under the same base
Hello, Punch!
Solve for $\displaystyle x\!:\;\;\log_5(x+7) - \log_{\sqrt{5}}(x+1)\:=\:1$ .[1]
$\displaystyle \text{Let }\log_{\sqrt{5}}(x+1) \:=\:P$ .[2]
$\displaystyle \text{Then: }\:(\sqrt{5})^P \:=\:x+1 \quad\Rightarrow\quad \left(5^{\frac{1}{2}}\right)^P \:=\:x+1 \quad\Rightarrow\quad 5^{\frac{P}{2}} \:=\:x+1 $
$\displaystyle \text{Take logs (base 5): }\;\log_5\left(5^{\frac{P}{2}}\right) \:=\:\log_5(x+1)$
$\displaystyle \text{Then: }\;\frac{P}{2}\cdot\underbrace{\log_5(5)}_{\text{T his is 1}} \:=\:\log_5(x+1) \quad\Rightarrow\quad \frac{P}{2} \:=\:\log_5(x+1)$
$\displaystyle \text{Hence: }\:P \:=\:2\log_5(x+1) \:=\:\log_5(x+1)^2$ .[3]
$\displaystyle \text{Equate {\color{blue}[2]} and {\color{blue}[3]}: }\;\log_{\sqrt{5}}(x+1) \;=\;\log_5(x+1)^2$
Substitute into [1]: .$\displaystyle \log_5(x+7) - \log_5(x+1)^2 \;=\;1 \quad\Rightarrow\quad \log_5\bigg[\frac{x+7}{(x+1)^2}\bigg] \;=\;1$
Exponentiate both sides: .$\displaystyle \frac{x+7}{(x+1)^2} \:=\:5^1$
Simplify: .$\displaystyle x+7 \:=\:5(x+1)^2 \quad\Rightarrow\quad 5x^2 + 9x - 2 \:=\:0$
Factor: .$\displaystyle (x+2)(5x-1) \:=\:0$
And we have two roots: .$\displaystyle x \;=\;-2,\:\tfrac{1}{5}$
But $\displaystyle -2$ is an extraneous root.
The solution is: .$\displaystyle \boxed{x \:=\:\frac{1}{5}}$