what is wrong with this natural log conversion why can't we convert this: $$4e^{1+3x}-9e^{5-2x}$$
to this: $$(1+3x)\ln(4)-(5-2x)\ln(9)$$
or this: $$(1+3x)\ln(4)+(5-2x)\ln(-9)$$
this comes from the q, solve: $$4e^{1+3x}-9e^{5-2x}=0$$
do the above conversions not work because of the ln of $0$? (that's why I posed it first without the $= 0$) or independently of that issue?
 A: The equation
$$
4e^{1+3x}-9e^{5-2x}=0
$$
is equivalent to
$$
4e^{1+3x}=9e^{5-2x}
$$
and here you can take the logarithm of both sides:
$$
\log(4e^{1+3x})=\log(9e^{5-2x})
$$
that becomes
$$
\log4+\log e^{1+3x}=\log9+\log e^{5-2x}
$$
or
$$
\log 4+1+3x=\log 9+5-2x
$$
that's a linear equation.
None of your attempts is good, I'm afraid: for instance, in general,
$$
\log(a+b)\ne\log a +\log b
$$
because $\log a+\log b=\log(ab)$. The fact that $\log0$ doesn't exist is another problem, of course, but not the main one.
A: It is not correct that $4e^{1+3x}-9e^{5-2x}$ is equal to $\ln(4e^{1+3x}) - \ln(9^{5-2x})$, nor that $\ln(4e^{1+3x}-9e^{5-2x})$ is equal to $\ln(4e^{1+3x}) - \ln(9^{5-2x})$. The first is wrong because numbers are not equal to their logarithms.  The second is wrong because $\ln(A-B)$ is not $\ln A-\ln B$.
But we can say that if $4e^{1+3x}-9e^{5-2x}=0$ then $\ln(4e^{1+3x})-\ln(9e^{5-2x})=0$.  The reason for that is that if $4e^{1+3x}-9e^{5-2x}=0$ then $4e^{1+3x}=9e^{5-2x}$.  If these two expressions are both the same number, then they both have the same logarithm, so $\ln(4e^{1+3x})=\ln(9e^{5-2x})$.  And from there one gets $(\ln4)+(1+3x)=(\ln9)+(5-2x)$, etc., and that can be quickly solved for $x$.
A: Remember that $\log(ab)= \log(a) + \log(b)$! So then could you then do this:
$$4e^{1+3x}-9e^{5-2x} = 0 $$
$$\ln(4) + 1 + 3x -1(\ln(9) + 5 -2x) = 0$$
seems like this gives me the same answer as you got but I am still not sure because I haven't taken the ln of the right hand side and it also seems to violate log(a + b) ≠ log(a) + log (b)
