Found this little puzzle on facebook, not sure if it was a joke.
Find $x$ $$\frac{(3x^2-27)(8x^2)^6}{4(9-3x)(x^2+3x)}=\frac{\tan(x+4)}{\log(x+\frac{1}{4})}$$
I'm thinking
- LHS numerator: taking the factor 3 out and expanding the right bracket
- LHS denominator: taking the factor of -3 from the left bracket and x from the right bracket
$$\frac{3(x^2-9)(8^6x^{12})}{-12x(x-3)(x+3)}=\frac{tan(x+4)}{log(x+\frac{1}{4})} $$
$$\frac{(x^2-9)(8^6x^{12})}{-4x(x^2-9)}=\frac{tan(x+4)}{log(x+\frac{1}{4})} \quad \text{taking the factor of 3 out and expanding noting the identity} $$
$$\frac{(8.8^5x^{12})}{-4x}=\frac{tan(x+4)}{log(x+\frac{1}{4})} \quad \text{cancelling of course, rearranging $8^6 = 8.8^5$} $$ $$(-2.8^5x^{11})=\frac{tan(x+4)}{log(x+\frac{1}{4})} \quad \text{even more cancelling} $$
I get this far not sure what happens next thanks
Perhaps
$$(-2.8^5)=\frac{tan(x+4)}{x^{11}log(x+\frac{1}{4})} $$
And then....
$$(-2.8^5)=\frac{tan(x+4)}{log\left(x+\frac{1}{4}\right)^{x^{11}}} $$
No how about
$$(-2.8^5x^{11})=\frac{tan(x+4)}{log(x+\frac{1}{4})} $$
$$log\left(x+\frac{1}{4}\right)(-2.8^5x^{11})=tan(x+4) $$
Now think of a triangle $tan(A) = \frac{Opposite}{Adjacent}$
So.... $$\frac{log\left(x+\frac{1}{4}\right)(-2.8^5x^{11})}{1}=tan(x+4) \quad \text{opp: $log\left(x+\frac{1}{4}\right)(-2.8^5x^{11})$, adj: 1} $$
So.... $$hypotenuse = \sqrt{1^2 + \left(log\left(x+\frac{1}{4}\right)(-2.8^5x^{11})\right)^2} $$
$$hypotenuse = \sqrt{1 + 2log\left(x+\frac{1}{4}\right)(4.8^{10}x^{22})} $$ $$hypotenuse = \sqrt{1 + log\left(x+\frac{1}{4}\right)(8.8^{10}x^{22})} $$ $$hypotenuse = \sqrt{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})} $$
$$cos(x+4) = A/H $$ $$cos(x+4) = \frac{1}{\sqrt{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})}}$$ $$cos^2(x+4) = \frac{1}{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})}$$ $$1-sin^2(x+4) = \frac{1}{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})}, \quad \text{using the idea that $cos^2(x) = 1 - sin^2(x) $}$$
$$1-\frac{log\left(x+\frac{1}{4}\right)(-2.8^5x^{11})}{\sqrt{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})}}\frac{log\left(x+\frac{1}{4}\right)(-2.8^5x^{11})}{\sqrt{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})}} = \frac{1}{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})}$$
$$1-\frac{\left(log\left(x+\frac{1}{4}\right)(-2.8^5x^{11}) \right)^2}{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})} = \frac{1}{1 + log\left(x+\frac{1}{4}\right)(8^{11}x^{22})}$$
$$(1 - log\left(x+\frac{1}{4}\right)(8^{11}x^{22}))+log\left(x+\frac{1}{4}\right)(8^{11}x^{22}) = 1$$
$$-( log\left(x+\frac{1}{4}\right)(8^{11}x^{22}))+log\left(x+\frac{1}{4}\right)(8^{11}x^{22}) = 0$$
$$0 = 0 $$
dammit.....
It was to big to fit into a comment ;)
$$\frac{(3x^2-27)(8x^2)^6}{4(9-3x)(x^2+3x)}=\frac{\tan(x+4)}{\ln\left(x+\frac{1}{4}\right)}\Longleftrightarrow$$ $$\frac{(3x^2-27)(8x^2)^6}{4x(x+3)(9-3x)}=\frac{\tan(x+4)}{\ln\left(x+\frac{1}{4}\right)}\Longleftrightarrow$$ $$\frac{(3x^2-27)(8x^2)^6}{12x(3-x)(x+3)}=\frac{\tan(x+4)}{\ln\left(x+\frac{1}{4}\right)}\Longleftrightarrow$$ $$\frac{3(x^2-9)(8x^2)^6}{12x(3-x)(x+3)}=\frac{\tan(x+4)}{\ln\left(x+\frac{1}{4}\right)}\Longleftrightarrow$$ $$\frac{3(8x^2)^6\left(x^2-3^2\right)}{12x(3-x)(x+3)}=\frac{\tan(x+4)}{\ln\left(x+\frac{1}{4}\right)}\Longleftrightarrow$$ $$\frac{3(x-3)(x+3)(8x^2)^6}{12x(3-x)(x+3)}=\frac{\tan(x+4)}{\ln\left(x+\frac{1}{4}\right)}\Longleftrightarrow$$ $$\frac{786432x^{12}(x-3)(x+3)}{12x(3-x)(x+3)}=\frac{\tan(x+4)}{\ln\left(x+\frac{1}{4}\right)}\Longleftrightarrow$$
So $\frac{786432x^{12}(x-3)(x+3)}{12x(3-x)(x+3)}=-65536x^{11}$ if $x\ne -3,x\ne 0,x\ne 3$:
$$-65536x^{11}=\frac{\tan(x+4)}{\ln\left(x+\frac{1}{4}\right)}\Longleftrightarrow$$ $$\frac{\frac{\tan(x+4)}{\ln\left(x+\frac{1}{4}\right)}}{x^{11}}=-65536\Longleftrightarrow$$ $$\frac{\tan(x+4)}{x^{11}\ln\left(x+\frac{1}{4}\right)}=-65536\Longleftrightarrow$$ $$\frac{\tan(x+4)}{x^{11}\ln\left(x+\frac{1}{4}\right)}+65536=0\Longleftrightarrow$$ $$65536x^{11}\ln\left(x+\frac{1}{4}\right)+\tan(x+4)=0\Longleftrightarrow$$ $$\left(\sec(x+4)\right)\left(65536x^{11}\ln\left(x+\frac{1}{4}\right)\cos(x+4)+\sin(x+4)\right)=0\Longleftrightarrow$$ $$\sec(x+4)=0\space\vee\space 65536x^{11}\ln\left(x+\frac{1}{4}\right)\cos(x+4)+\sin(x+4)=0$$