# Thread: Logarithm problem (sorry if this in the wrong section!)

1. ## Logarithm problem (sorry if this in the wrong section!)

I couldn't seem to find a board on logarithms so apologised if this is in the incorrect section. Apologies for formatting too.

I'm trying to work out log 4^2x-1 = log 3^x+1

I've done this so far (but I think I've made it more difficult than it needs to be)

i) (2x – 1)log 4 = (x + 1)log 3
ii) 2x-1= (x+1)log⁡3 / log⁡4
iii) 2x - (x+1)log 3 / log 4 = 1
but I do not know where to go from there...and like I said I think there is an easier more straightforward way to do this.

I'd appreciate any help. Thank you.

2. $\displaystyle \begin{gathered} \left( {2x - 1} \right)\log 4 = \left( {x + 1} \right)\log 3 \hfill \\ \leftrightarrow \left( {2\log 4 - \log 3} \right)x = \log 3 + \log 4 \hfill \\ \leftrightarrow x = \frac{{\log 12}} {{\log \frac{{16}} {3}}} \hfill \\ \end{gathered}$

3. Hello, Jamie!

Solve: . $\displaystyle \log\left(4^{2x-1}\right) \: = \:\log\left(3^{x+1}\right)$

I've done this so far . . .

$\displaystyle (2x – 1)\log 4 \:= \x + 1)\log 3$ . . . . Right!
We want to solve for $\displaystyle x$, right?

Expand: . $\displaystyle 2x\!\cdot\!\log 4 - \log4 \;=\;x\!\cdot\!\log3 + \log 3$

Get the $\displaystyle x$'s together: . $\displaystyle 2x\!\cdot\log4 - x\!\cdot\!\log3 \;=\;\log 3 + \log 4$

Factor: . $\displaystyle x\left(2\log4 - \log 3\right) \:=\:\log 3 + \log 4$

Therefore: . $\displaystyle \boxed{x \;=\;\frac{\log3 + \log4}{2\log4 - \log3}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Peritus simplified it further, using logarithm rules.

. . $\displaystyle \log3 + \log4 \;=\;\log(3\cdot4) \;=\;\log12$

. . $\displaystyle 2\log4 - \log 3 \;=\;\log(4^2) - \log3 \;=\;\log16 - \log3 \;=\;\log\left(\frac{16}{3}\right)$

And we have: . $\displaystyle \frac{\log(12)}{\log\left(\frac{16}{3}\right)} \;=\;\log_{\frac{16}{3}}(12)$

I've always wanted to do that!
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