1. ## Logarithms!

$\displaystyle log_5(x+7) - log_(\sqrt5) (x+1)=1$

$\displaystyle \log_5(x+7) - \log_5\frac{1}{2}(x+1) = 1$

$\displaystyle \log_5\left(\frac{x+7}{\frac{1}{2}(x+1)}\right) = 1$

$\displaystyle \frac{x+7}{\frac{1}{2}(x+1)} = 5$

Can you take it from here?

3. 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

4. Originally Posted by Punch
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 $\displaystyle log_5(x+7) - log_(\sqrt5) (x+1)$

The log having a base of square root is that making it difficult for me
consider the relation between $\displaystyle log_2(x)$ and $\displaystyle log_4(y)$, how are x and y related?

5. Sorry but I don'tunderstand

6. $\displaystyle log_5(x+7) - log_(\sqrt5) (x+1)=1$

but i just need to solve this equation

7. Originally Posted by Punch
$\displaystyle log_5(x+7) - log_(\sqrt5) (x+1)=1$

but i just need to solve this equation
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

8. GAWD, So it just means that I have to square both \sqrt5 and (x+1)

9. I got it, I got it yay thanks a lot!

10. 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: .$\displaystyle \boxed{x \:=\:\frac{1}{5}}$

11. thanks but what's an extraneous root?

12. Originally Posted by Punch
thanks but what's an extraneous root?
try to plug -2 in your equation and see what happens.

13. lol understood thanks all for being so helpful to this maths noob here