I'm a little stuck (again).
How would I go about solving an equation like this:
$\displaystyle 4^2^x-4^x-20=0$
And a more complicated one such as:
$\displaystyle 2^x+12(2)^-^x=7$
All help greatly appreciated!
This is a quadratic in 4^x and happily it will factor to $\displaystyle (4^x-5)(4^x+4)=0$. Note that $\displaystyle 4^x > 0$
Multiply through by the LCD which is 2^x since $\displaystyle 2^{-x} = \dfrac{1}{2^x}$. You'll get $\displaystyle 2^{2x} + 12 = 7 \cdot 2^x$ which, again, is a quadratic equationAnd a more complicated one such as:
$\displaystyle 2^x+12(2)^-^x=7$
All help greatly appreciated!
Thanks! However, I'm a little stuck once we get to $\displaystyle 7 \cdot 2^x$. The answer in the back of the book shows $\displaystyle x=2$ and $\displaystyle x=\dfrac{log3}{log2}$.
How would we multiply $\displaystyle 7$ with $\displaystyle 2^x$ but still come to such an answer?
Thanks!
You were told what to do:
This is not the answer, it is how to get the answer. And how to do this was explained to you with the first question. Let $\displaystyle 2^x = w$. Re-arrange to get a quadratic equation in w that is equal to zero. Solve for w. Hence solve for x.
Please make an atttempt. Show your work, say where you get stuck if you need more help.