Solve the equation
4^(2x)-4^(x-1)-12=0
How is this equation solved using logs?? Please help.
4^(2x) -4^(x-1) -12 = 0
Cannot take the logs of both sides because log(0) is undefined or there are no log(0). So,
4^(2x) -4^(x-1) = 12
Take the logs of both sides,
2x*log(4) -(x-1)log(4) = log(12)
2x*log(4) -x*log(4) +log(4) = log(4*3)
x*log(4) +log(4) = log(4) +log(3)
x*log(4) = log(3)
x = log(3) / log(4) --------------answer.
Or, x = (1/2)[log(3) / log(2)]
Hello, haku!
Solve: .$\displaystyle 4^{2x} - 4^{x-1} - 12\:=\:0$
We have: .$\displaystyle 4^{2x} - 4^x\!\cdot 4^{-1} - 12 \;=\;0 \quad\Rightarrow\quad 4^{2x} - \frac{1}{4}\!\cdot\!4^x - 12 \;=\;0$
Multiply by 4: .$\displaystyle 4\cdot4^{2x} - 4^x - 48 \;=\;0$
We have a quadratic: . $\displaystyle 4(4^x)^2 - 4^x - 48 \;=\;0$
Quadratic Formula: . $\displaystyle 4^x\;=\;\frac{1 +\sqrt{769}}{8}$
Then: .$\displaystyle x \:=\:\log_4\left(\frac{1 + \sqrt{769}}{8}\right) \;=\;\frac{\ln\left(\frac{1+\sqrt{769}}{8}\right)} {\ln(4)} \;\approx\;0.922264366$
Yes.
[I just took a peek. I'm home for a while.]
Log of the original LHS is not as I wrte it. I fell on the trap.
log(a +b) is not log(a) +log(b), as I warned in another reply before.
And, normally I check my answer against the original equation, quickly or in "scratch paper" or calculator, before I say that is my answer. I forgot to do that this morning before I left for work.