How can i solve this $4^{2+x} + 15 * 4^{x^2}-16^{x^2-\frac{x}{2}}=0$?

Question

Can you help me to solve this equation?

$$4^{2+x} + 15 * 4^{x^2}-16^{x^2-\frac{x}{2}}=0$$

See also : https://math.stackexchange.com/questions/384090/find-all-real-numbers-x-for-which-frac8x27x12x18x-frac76 — lab bhattacharjee, Dec 17 '17 at 15:38

score 1 · Accepted Answer · answered Dec 17 '17 at 15:35

1

Hint:

$$0=16(4^x)+15(4^{x^2})-4^{2x^2-x}$$

$$\iff16(4^x)^2+15(4^{x^2})(4^x)-(4^{x^2})^2=0$$

Divide both sides by $(4^{x^2})^2\ne0$ to find $$16\left(4^{x-x^2}\right)^2+15\left(4^{x-x^2}\right)-1=0$$

Observer that $16a^2+15a-1=(16a-1)(a+1)$

Now for $x-x^2,4^{x-x^2}>0$

answered Dec 17 '17 at 15:35

lab bhattacharjee

1 Answers1