<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] Notice, $$\ln(2x^2+9x-5)=1+\ln(x^2+2x-15)$$
$$\ln(2x^2+9x-5)=\ln e+\ln(x^2+2x-15)$$
$$\ln(2x^2+9x-5)=\ln(e(x^2+2x-15))$$
$$2x^2+9x-5=e(x^2+2x-15)$$
$$(e-2)x^2-(9-2e)x-5(3e-1)=0$$
$$\iff x=\frac{(9-2e)\pm\sqrt{(9-2e)^2+20(e-2)(3e-1)}}{2(e-2)}$$
$$=\frac{(9-2e)\pm(8e-11)}{2(e-2)}$$
$$x=\frac{3e-1}{e-2}$$
or $$x=\frac{10-5e}{e-2}=-5$$ but for $x=-5$ log is undefined thus the above value is unacceptable 
Hence, the correct value of $x$ 
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{x=\frac{3e-1}{e-2}}}$$
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