So, please don't ignore me cause I'm only going to give you the equation but I've tried several different routes of attacking this equation and I can't seem to figure it out!
4^x + 6(4^(-x)) = 5
Hello, Mike!
Solve for $\displaystyle x\!:\;\;4^x + 6\cdot4^{-x} \:=\: 5$
Multiply by $\displaystyle 4^x\!:\;\;4^{2x} + 6 \:=\:5\cdot4^x\quad\Rightarrow\quad 4^{2x} - 5\cdot 4^x + 6 \:=\:0$
Factor: .$\displaystyle \left(4^x-2\right)\left(4^x - 3\right) \:=\:0$
Then we have:
. . $\displaystyle 4^x - 2 \:=\:0 \quad\Rightarrow\quad 4^x \:=\:2 \quad\Rightarrow\quad x \:=\:\log_4(2) \quad\Rightarrow\quad\boxed{ x \:=\:\frac{1}{2}}$
. . $\displaystyle 4^x - 3 \:=\:0\quad\Rightarrow\quad 4^x \:=\:3 \quad\Rightarrow\quad x \:=\:\log_4(3)\quad\Rightarrow\quad\boxed{ x \:=\:\frac{\ln(3)}{\ln(4)}} $
Let me know if I'm making a mistae by just reposting in this thread.... It's the exact same question, so I didn't think it was worthy of a thread wit hthe same name.
Here's the question:
log(3x-5)base4 = log(11)base4 + log(2)base4
I got it to
log(3x-5)base4 = log(22)base4