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)log3 / log4
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.
Hello, Jamie!
We want to solve for , right?Solve: .
I've done this so far . . .
x + 1)\log 3" alt=" (2x – 1)\log 4 \:= \x + 1)\log 3" /> . . . . Right!
Expand: .
Get the 's together: .
Factor: .
Therefore: .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Peritus simplified it further, using logarithm rules.
. .
. .
And we have: .
I've always wanted to do that!
.