1. ## Logarithms

Hi,

$\text{If} \log 4 + 2 \log p = 2, \,\,\text{find p.}$

Initially I tried this:

$\log\, 4 + \log p^2 = 2$

$\log\, 4p^2 = \log 2\bigg]\Rightarrow$ logs cancel since they're of the same base

$4p^2 = 2$

$p = \sqrt{\frac{1}{2}}$

I found my answer to be incorrect. I want to know why.

2. Originally Posted by Hellbent
Hi,

$\text{If} \log 4 + 2 \log p = 2, \,\,\text{find p.}$

Initially I tried this:

$\log\, 4 + \log p^2 = 2$

$\log\, 4p^2 = \log 2\bigg]\Rightarrow$ logs cancel since they're of the same base
Why did you introduce a log on the right hand side? There shouldn't be one there.

Your third line should read $\log(4p^2) = 2$

So you get $4p^2 = 10^2$ in base 10 or $4p^2 = e^2$ in base e

edit: you touched upon it but remember to discard the negative value of p since $p>0$

edit 2: I get an answer of $p=5$

3. This line is not true.

$\log\, 4p^2 = \log 2$

This is true:

$\log\, 4p^2 = 2$

4. Spoiler:
$\log{4} + 2\log p = 2 \Rightarrow \log{2^2} + 2\log{p} = 2 \Rightarrow 2\log{2} + 2\log{p} = 2 \Rightarrow 2\left(\log{2}+\log{p}\right) = 2$.
Spoiler:
So $\log{2}+\log{p} = 1$, therefore $\log{p} = 1-\log{2}$, just put $p$ on one side.
Spoiler:
$\displaystyle \log{p} = 1-\log{2} \Leftrightarrow e^{\log{p}} = e^{1-\log{2}} \Rightarrow p = e \left(e^{-\log(2)\right) = \frac{e}{e^{\log{2}}} = \frac{e}{2}.$