Find $x$ such that
$\dfrac{6(3x^2-27)8x^2}{4(9-3x)(x^2+3x)} = \dfrac{\tan(x+4)}{\ln(x+\frac{1}{4})}$
I've managed to reduce the left hand side so that
$-12x= \dfrac{\tan(x+4)}{\ln(x+\frac{1}{4})}= \dfrac{\tan(x+4)}{\ln(\frac{4x+1}{4})} = \dfrac{\tan(x+4)}{\ln(4x+1)-\ln(4)}$
But I'm unsure on how to deal with both the $\tan(x+4)$ and $\ln(4x+1)$. I've tried using the maclaurin series expansion for $\tan(\alpha)$ and $\ln(1+\beta)$, with $\alpha=x+4$ and $\beta=4x$ but this just gives an over complicated fraction which I cannot reduce to get a solution for $x$.