1. Solve the equation

Solve the equation
4^(2x)-4^(x-1)-12=0

2. Originally Posted by haku
Solve the equation
4^(2x)-4^(x-1)-12=0

Rewrite this as:

$\displaystyle 4^{2x}-4^{x-1}-12 = (4^x)^2 - (4^x)/4 -12=0$

Then change the variable to $\displaystyle y=4^x$, so that $\displaystyle x=\log_4(y)$. Then:

$\displaystyle y^2 - y/4-12=0$

Now solve for $\displaystyle y$, then for $\displaystyle x$

RonL

3. 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)]

4. Thanks for the help guys!

5. 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: .$\displaystyle x \:=\:\log_4\left(\frac{1 + \sqrt{769}}{8}\right) \;=\;\frac{\ln\left(\frac{1+\sqrt{769}}{8}\right)} {\ln(4)} \;\approx\;0.922264366$

6. Originally Posted by ticbol
4^(2x) -4^(x-1) = 12

Take the logs of both sides,
2x*log(4) -(x-1)log(4) = log(12)