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 can take $e$ to both sides in the beginning, but the simplification on the left side will be $e^{\ln(2x) + \ln 5} = e^{\ln(2x)} \cdot e^{\ln 5} = 2x \cdot 5 = 10x$. – user307169 Jun 29 '17 at 12:12
• If you're gonna write here I'd suggest that you start learning MathJAX so you can format your mathematical expressions better. A guide can be found here: (math.meta.stackexchange.com/questions/5020/…). Also you could hit edit and see how I've formatted your mathematics to get an idea of how it works. – skyking Jun 29 '17 at 12:13

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.

• Why is it that we have e^(ln(2x)+ln(5)) and not e^ln(2x) + e^ln(5) seperately? If I have a simple equation like: 1/2x^2 + 1/2x = 1/2, and I want to simplify by manipulating the equation by doing everything times 2. Then I will have to multiply every part of the equation by 2 right? => 2(1/2x^2) + 2(1/2x) = 2(1/2). – Hikato Jun 29 '17 at 12:19
• You are correct with your multiplying-by-2 example. This is because "multiplication distributes over addition". That is, $a(b+c) = ab + ac$. However, exponentiation does not distribute over addition. That is, we do not have $a^{b+c} = a^b + a^c$ in general. You can either memorize it as a rule of algebra, or you will need to have to dig deeper into what exponentiation means to figure out why this is so. – Bob Krueger Jun 29 '17 at 14:38

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.

• So basically if you want to take e^x from any side of an equation, you have to include everything that is on that side in e^(here)? – Hikato Jun 29 '17 at 12:28
• @David Well, yeah. That's not specific to $e^.$ as well. If I tell you that $x$ is the same thing as $y$, then clearly, you can conclude that $f(x)=f(y)$. But not all functions then satisfy the property $f(a+b)=f(a)+f(b)$. The exponential funciton, for example, does not. – 5xum Jun 29 '17 at 12:30
• Thanks a lot, that makes sense :). I have one more question, why is it that when I put Y1 = ln(8-x^2) - ln(3-x) in the calculator and Y2 = ln((8-x^2)/(3-x)), that they differ? They are almost the same but Y2 has more solutions somehow when I graph it. Y1 and Y2 are the same right? – Hikato Jun 29 '17 at 12:58
• @David is that $8$ supposed to be a $9$? – 5xum Jun 29 '17 at 13:00
• @David Oh, yeah, I was wrong. The thing is that $\ln(8-x^2)-\ln(3-x)$ is defined only when both terms inside $\ln$ are positive, while $\ln(\frac{8-x^2}{3-x})$ is defined when the entire fraction is negative. So, for $x=4$, $\ln(8-x^2)-\ln(3-x)$ is not defined because $\ln(-8)$ and $\ln(-1)$ are not defined, but $\ln(\frac{8-x^2}{3-x})=\ln\frac{8}{1}$ is defined. – 5xum Jun 29 '17 at 13:30