Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
Example:
$$\ln(2x) + \ln(5) = 0$$
To solve for x, use the ln property: $\ln(2x) + \ln(5) = \ln(10x)$
$$\begin{aligned}\ln(10x) &= 0\\ e^{\ln(10x)} &= e^0\\ 10x &= 1\\ x &= \frac{1}{10}\end {aligned}$$
I wonder why you can't do: $e^{\ln(2x)} + e^{\ln(5)} = e^0 \implies 2x + 5 = 1$. Which is another outcome, but incorrect.
Why do you have to use the ln property to add up $\ln(2x)$ and $\ln(5)$ first before continuing the equation? Why can't you take the $e^x$ from those right away?
Thank you.
You are erroneously supposing that $e^{x+y} = e^x + e^y$ (take for example $1 = e^0 = e^{1+(-1)}\ne e^1 + e^{-1}\approx 3.1$).
That is, just because $\ln(2x) + \ln(5) = 0$, we surely have $e^{\ln(2x) + \ln(5)} = e^0 = 1$, but we don't then have $e^{\ln(2x)} + e^{\ln(5)} = 1$.
The correct step is to use $e^{x+y} = e^x e^y$, so addition in the exponent turns into multiplication. This gives $e^{\ln(2x)} e^{\ln(5)} = 1$, or $2x \cdot 5 = 1$, as you did with the correct approach.
You start with an equation $$A + B = C$$
What you can do is change that into $$e^{A+B} = e^C$$ and that leads to the correct solution, since $$e^{\ln(2x) + \ln(5)} = e^{\ln 2x}\cdot e^{\ln(5)} = 10 x$$
What you cannot do is change that into $$e^A + e^B = e^C$$
because $$e^{A+B}\neq e^A+e^B$$ in general.
