3
$\begingroup$

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.

$\endgroup$
2
  • 3
    $\begingroup$ You can take $e$ to both sides in the beginning, but the simplification on the left side will be $e^{\ln(2x) + \ln 5} = e^{\ln(2x)} \cdot e^{\ln 5} = 2x \cdot 5 = 10x$. $\endgroup$ – user307169 Jun 29 '17 at 12:12
  • 1
    $\begingroup$ 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. $\endgroup$ – skyking Jun 29 '17 at 12:13
13
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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). $\endgroup$ – Hikato Jun 29 '17 at 12:19
  • 4
    $\begingroup$ 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. $\endgroup$ – Bob Krueger Jun 29 '17 at 14:38
7
$\begingroup$

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.

$\endgroup$
9
  • $\begingroup$ 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)? $\endgroup$ – Hikato Jun 29 '17 at 12:28
  • 1
    $\begingroup$ @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. $\endgroup$ – 5xum Jun 29 '17 at 12:30
  • $\begingroup$ 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? $\endgroup$ – Hikato Jun 29 '17 at 12:58
  • $\begingroup$ @David is that $8$ supposed to be a $9$? $\endgroup$ – 5xum Jun 29 '17 at 13:00
  • 1
    $\begingroup$ @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. $\endgroup$ – 5xum Jun 29 '17 at 13:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.