Solving $2^{2x+1} - 2^{x+4} = 2^3 - 2^x$

Question

$$2^{2x+1} - 2^{x+4} = 2^3 - 2^x$$

How can I solve an exponential equation that has many terms as the one above. Include more than one method if available.

Answer 1

The key here is to transform this into a quadratic equation.

$$2^{2x+1} - 2^{x+4} = 2^3 - 2^x$$ can be rewritten as $$2 \cdot 2^{2x} - 2^4 \cdot 2^x = 2^3 -2^x.$$

Temporarily we will express $2^x=u$ and we find that the equation turns into $$2u^2 - 16u = 8 - u.$$ Finally writing this as $$2u^2 - 15u - 8 = 0$$ we can factor or use the quadratic equation to solve for $u$. Then remembering that $u=2^x$ we can find the solution for $x$ by writing $\log_2(u)=\ln(u)/\ln(2)=x$. Just be sure to try plugging the answers back into the original equation at the end.

Answer 2

$$\Longleftrightarrow 2^x(2^{x+1}+1)=2^3(2^{x+1}+1)$$

Answer 3

Factor out $2^{x+1}+1$ from both sides, you will get: $$2^x(2^{x+1}+1)=2^3(2^{x+1}+1)$$ One solution will be $2^{x+1}+1=0$ and second will be $2^x=2^3$. First equation does not have real solutions, but over complex number it have a solution $$x=-1+\frac{i\pi}{\ln2}+\frac{2i\pi k}{\ln2},k\in\mathbb{Z}$$ Second equation have only one real solution $x=3$ and infinitely many complex solutions $$x=3+\frac{2i\pi k}{\ln2},k\in\mathbb{Z}$$

Answer 4

Given that the bases are the same, correct me if I'm wrong, but isn't your equation:

$$2^{2x+1} - 2^{x+4} = 2^3 - 2^x$$

equivalent to this?

$$(2x+1) - (x+4) = 3 - x$$

in which case it's easy to show $x = 3$? Plugging 3 into your original equation shows that it is indeed the answer.

Answer 5

Hint $\ ac = bd\,\Rightarrow\, b(a\!-\!b\!-\!c\!+\!d) = (a\!-\!b)(b\!+\!c).\ $ Let $\ a,b,c,d\, =\, 2^{2x+1},2^{x+4},2^3,2^x$