Encontrar $x$ $\frac{(3x^2-27)(8x^2)^6}{4(9-3x)(x^2+3x)}=\frac{\tan(x+4)}{\log(x+\frac{1}{4})}$

Question

Encuentra este pequeño puzzle en facebook, no se si era una broma.

Encontrar $x$ $$\frac{(3x^2-27)(8x^2)^6}{4(9-3x)(x^2+3x)}=\frac{\tan(x+4)}{\log(x+\frac{1}{4})}$$

Estoy pensando

• LHS numerador: tomando el factor 3 y ampliar el soporte derecho
• LHS denominador: tomando el factor de -3 de la izquierda y soporte de x desde el soporte derecho

$$\frac{3(x^2-9)(8^6x^{12})}{-12x(x-3)(x+3)}=\frac{tan(x+4)}{log(x+\frac{1}{4})} $$

$$\frac{(x^2-9)(8^6x^{12})}{-4x(x^2-9)}=\frac{tan(x+4)}{log(x+\frac{1}{4})} \quad \text{taking the factor of 3 out and expanding noting the identity} $$

$$\frac{(8.8^5x^{12})}{-4x}=\frac{tan(x+4)}{log(x+\frac{1}{4})} \quad \text{cancelling of course, rearranging $8^6 = 8.8^5$} $$ $$(-2.8^5x^{11})=\frac{tan(x+4)}{log(x+\frac{1}{4})} \quad \text{even more cancelling} $$

Yo llego hasta este punto no está seguro de lo que sucede a continuación gracias

---

Tal vez

$$(-2.8^5)=\frac{tan(x+4)}{x^{11}log(x+\frac{1}{4})} $$

Y entonces....

$$(-2.8^5)=\frac{tan(x+4)}{log\left(x+\frac{1}{4}\right)^{x^{11}}} $$

---

No ¿

$$(-2.8^5x^{11})=\frac{tan(x+4)}{log(x+\frac{1}{4})} $$

$$log\left(x+\frac{1}{4}\right)(-2.8^5x^{11})=tan(x+4) $$

Ahora pensar en un triángulo $tan(A) = \frac{Opposite}{Adjacent}$

Así que.... $$\frac{log\left(x+\frac{1}{4}\right)(-2.8^5x^{11})}{1}=tan(x+4) \quad \text{opp: $registro\left(x+\frac{1}{4}\right)(-2.8^5x^{11})$, adj: 1} $$

Así que.... $$hypotenuse = \sqrt{1^2 + \left(log\left(x+\frac{1}{4}\right)(-2.8^5x^{11})\right)^2} $$

$$hypotenuse = \sqrt{1 + 2log\left(x+\frac{1}{4}\right)(4.8^{10}x^{22})} $$ $$hypotenuse = \sqrt{1 + log\left(x+\frac{1}{4}\right)(8.8^{10}x^{22})} $$ $$hypotenuse = \sqrt{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})} $$

$$cos(x+4) = A/H $$ $$cos(x+4) = \frac{1}{\sqrt{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})}}$$ $$cos^2(x+4) = \frac{1}{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})}$$ $$1-sin^2(x+4) = \frac{1}{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})}, \quad \text{using the idea that $cos^2(x) = 1 - sen^2(x) $}$$

$$1-\frac{log\left(x+\frac{1}{4}\right)(-2.8^5x^{11})}{\sqrt{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})}}\frac{log\left(x+\frac{1}{4}\right)(-2.8^5x^{11})}{\sqrt{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})}} = \frac{1}{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})}$$

$$1-\frac{\left(log\left(x+\frac{1}{4}\right)(-2.8^5x^{11}) \right)^2}{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})} = \frac{1}{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})}$$

$$(1 - log\left(x+\frac{1}{4}\right)(8^{11}x^{22}))+log\left(x+\frac{1}{4}\right)(8^{11}x^{22}) = 1$$

$$-( log\left(x+\frac{1}{4}\right)(8^{11}x^{22}))+log\left(x+\frac{1}{4}\right)(8^{11}x^{22}) = 0$$

$$0 = 0 $$

maldita sea.....

Answer 1

Era grande para caber en un comentario ;)

$$\frac{(3x^2-27)(8x^2)^6}{4(9-3x)(x^2+3x)}=\frac{\tan(x+4)}{\ln\left(x+\frac{1}{4}\right)}\Longleftrightarrow$$ $$\frac{(3x^2-27)(8x^2)^6}{4x(x+3)(9-3x)}=\frac{\tan(x+4)}{\ln\left(x+\frac{1}{4}\right)}\Longleftrightarrow$$ $$\frac{(3x^2-27)(8x^2)^6}{12x(3-x)(x+3)}=\frac{\tan(x+4)}{\ln\left(x+\frac{1}{4}\right)}\Longleftrightarrow$$ $$\frac{3(x^2-9)(8x^2)^6}{12x(3-x)(x+3)}=\frac{\tan(x+4)}{\ln\left(x+\frac{1}{4}\right)}\Longleftrightarrow$$ $$\frac{3(8x^2)^6\left(x^2-3^2\right)}{12x(3-x)(x+3)}=\frac{\tan(x+4)}{\ln\left(x+\frac{1}{4}\right)}\Longleftrightarrow$$ $$\frac{3(x-3)(x+3)(8x^2)^6}{12x(3-x)(x+3)}=\frac{\tan(x+4)}{\ln\left(x+\frac{1}{4}\right)}\Longleftrightarrow$$ $$\frac{786432x^{12}(x-3)(x+3)}{12x(3-x)(x+3)}=\frac{\tan(x+4)}{\ln\left(x+\frac{1}{4}\right)}\Longleftrightarrow$$

---

Por lo $\frac{786432x^{12}(x-3)(x+3)}{12x(3-x)(x+3)}=-65536x^{11}$ si $x\ne -3,x\ne 0,x\ne 3$:

---

$$-65536x^{11}=\frac{\tan(x+4)}{\ln\left(x+\frac{1}{4}\right)}\Longleftrightarrow$$ $$\frac{\frac{\tan(x+4)}{\ln\left(x+\frac{1}{4}\right)}}{x^{11}}=-65536\Longleftrightarrow$$ $$\frac{\tan(x+4)}{x^{11}\ln\left(x+\frac{1}{4}\right)}=-65536\Longleftrightarrow$$ $$\frac{\tan(x+4)}{x^{11}\ln\left(x+\frac{1}{4}\right)}+65536=0\Longleftrightarrow$$ $$65536x^{11}\ln\left(x+\frac{1}{4}\right)+\tan(x+4)=0\Longleftrightarrow$$ $$\left(\sec(x+4)\right)\left(65536x^{11}\ln\left(x+\frac{1}{4}\right)\cos(x+4)+\sin(x+4)\right)=0\Longleftrightarrow$$ $$\sec(x+4)=0\space\vee\space 65536x^{11}\ln\left(x+\frac{1}{4}\right)\cos(x+4)+\sin(x+4)=0$$

Answer 2

Cómo acerca de esto....

$$(-2.8^5x^{11})=\frac{tan(x+4)}{log(x+\frac{1}{4})} $$

$$log\left(x+\frac{1}{4}\right)(-2.8^5x^{11})=tan(x+4) $$

Ahora pensar en un triángulo $tan(A) = \frac{Opposite}{Adjacent}$

Así que.... $$\frac{log\left(x+\frac{1}{4}\right)(-2.8^5x^{11})}{1}=tan(x+4) \quad \text{opp: $registro\left(x+\frac{1}{4}\right)(-2.8^5x^{11})$, adj: 1} $$

Así que.... $$hypotenuse = \sqrt{1^2 + \left(log\left(x+\frac{1}{4}\right)(-2.8^5x^{11})\right)^2} $$

$$hypotenuse = \sqrt{1 + 2log\left(x+\frac{1}{4}\right)(4.8^{10}x^{22})} $$ $$hypotenuse = \sqrt{1 + log\left(x+\frac{1}{4}\right)(8.8^{10}x^{22})} $$ $$hypotenuse = \sqrt{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})} $$

$$cos(x+4) = A/H $$ $$cos(x+4) = \frac{1}{\sqrt{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})}}$$ $$cos^2(x+4) = \frac{1}{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})}$$ $$1-sin^2(x+4) = \frac{1}{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})}, \quad \text{using the idea that $cos^2(x) = 1 - sen^2(x) $}$$

$$1-\frac{log\left(x+\frac{1}{4}\right)(-2.8^5x^{11})}{\sqrt{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})}}\frac{log\left(x+\frac{1}{4}\right)(-2.8^5x^{11})}{\sqrt{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})}} = \frac{1}{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})}$$

$$1-\frac{\left(log\left(x+\frac{1}{4}\right)(-2.8^5x^{11}) \right)^2}{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})} = \frac{1}{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})}$$

$$(1 - log\left(x+\frac{1}{4}\right)(8^{11}x^{22}))+log\left(x+\frac{1}{4}\right)(8^{11}x^{22}) = 1$$

$$-( log\left(x+\frac{1}{4}\right)(8^{11}x^{22}))+log\left(x+\frac{1}{4}\right)(8^{11}x^{22}) = 0$$

$$0 = 0 $$

Answer 3

Definitivamente se ve como que no puede resolver esta ecuación analíticamente, wolfram probablemente da una aproximación numérica de la solución. También, tenga en cuenta que la interpretación geométrica es equivalente a la fórmula $$1 + tg^2(x) = \frac{1}{cos^2(x)}$$ También tenga en cuenta que$(log(x))^2 \ne 2log(x) $, por lo que su fórmula para la hipotenusa es incorrecta.