<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] No, it is equal to
$2x^2+9x−5 = e^{1+ln(x^2+2x−15)} = e(x^2+2x-15)$
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