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?
 A: 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 $x$ 
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{x=\frac{3e-1}{e-2}}}$$
A: No, it is equal to
$2x^2+9x−5 = e^{1+ln(x^2+2x−15)} = e(x^2+2x-15)$
A: $$\ln(2x^2+9x-5)=1+\ln(x^2+2x-15)$$
$$\ln(2x^2+9x-5)=\ln e+\ln(x^2+2x-15)$$
$$\ln(2x^2+9x-5)=\ln(e(x^2+2x-15))$$
$$2x^2+9x-5=e(x^2+2x-15)$$
$$(2-e)x^2+(9-2e)x-(5+15e)=0$$
that is quadratics with solution
$$x_1=3+\frac{5}{e-2}$$ and
$$x_2=-5$$ witch is unacceptable.
A: Zero point's first step is correct.  Alternatively, you could combine both natural logs to get
$$\ln\frac{2x^2+9x-5}{x^2+2x-15}=1$$
$$\frac{2x^2+9x-5}{x^2+2x-15}=e$$
It may help to factor next.
$$\frac{(2x-1)(x+5)}{(x+5)(x-3)}=e$$
Note that $x$ cannot equal $-5$ or else both natural logs in the original problem are not defined.  So we have
$$\frac{2x-1}{x-3}=e$$
Now we can solve for $x$.
$$2x-1=e(x-3)$$
$$(e-2)x=3e-1$$
$$x=\frac{3e-1}{e-2}$$
We are not out of the woods yet.  It still needs to be verified that both quadratics in the original problem are positive for the logarithms to be defined.  Based on the factorizations, that requires $x>3$ or $x<-5$.  We know $e>2$, so $x$ is positive.  For the first condition, we can rewrite $x$ as $3+\frac5{e-2}>3$, so the solution checks.
