<s>[INST] Given that $\ln (2x^2+9x-5)=1+\ln (x^2+2x-15)$ find $x$ in terms of $e$ Given that $\ln (2x^2+9x-5)=1+\ln (x^2+2x-15)$ find $x$ in terms of $e$. Then the left side is $2x^2+9x-5=\ln (1)+x^2+2x-15$, is that true? How should I proceed?
 [/INST] $$\ln(2x^2+9x-5)=1+\ln(x^2+2x-15)$$
$$\ln(2x^2+9x-5)=\ln e+\ln(x^2+2x-15)$$
$$\ln(2x^2+9x-5)=\ln(e(x^2+2x-15))$$
$$2x^2+9x-5=e(x^2+2x-15)$$
$$(2-e)x^2+(9-2e)x-(5+15e)=0$$
that is quadratics with solution
$$x_1=3+\frac{5}{e-2}$$ and
$$x_2=-5$$ witch is unacceptable.
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