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!

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- Jan 6th 2011, 01:40 PMyoungb11Solving an exponential equation
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! - Jan 6th 2011, 01:45 PMe^(i*pi)
This is a quadratic in 4^x and happily it will factor to $\displaystyle (4^x-5)(4^x+4)=0$. Note that $\displaystyle 4^x > 0$

Quote:

And a more complicated one such as:

$\displaystyle 2^x+12(2)^-^x=7$

All help greatly appreciated!

- Jan 6th 2011, 01:46 PMRandom Variable
$\displaystyle (4^{x})^{2} - 4^{x} - 20 = 0 $

now let $\displaystyle u = 4^{x} $ - Jan 6th 2011, 04:38 PMyoungb11
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! - Jan 6th 2011, 04:44 PMmr fantastic
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. - Jan 6th 2011, 04:46 PMyoungb11