How can we solve the following natural logarithms? I'm confused with this stuff:
- $\ln(x+1) - \ln x = \ln 3$
- $\ln(x+1) + \ln x = \ln 2$
2 Answers
$\begingroup$Here is a worked example slightly different to your question; solve the following for $q$: $$\ln(q+2)-\ln(q)=\ln(5)\tag{1}$$ Using the laws for adding/subtracting logarithms: $$\ln(a)-\ln(b)=\ln\left(\dfrac{a}{b}\right)$$ $$\ln(a)+\ln(b)=\ln(ab)$$ we can write $(1)$ as: $$\ln\left(\dfrac{q+2}{q}\right)=\ln(5).$$ this is the key step because once we have a single logarithm on both sides of the equation we can "invert" it to leave what is inside the logarithm by applying exponentiation to both sides (see Appendix). So, to undo the logarithm we take the exponential of both sides of the equation: $$e^{\ln\left(\dfrac{q+2}{q}\right)}=e^{\ln(5)}\tag{2}$$ and because $$e^{\ln(x)}=x$$ $(2)$ becomes: $$\dfrac{q+2}{q}=5$$ which can be rearranged to solve for $q$ and we find: $$q=\dfrac{1}{2}.$$Appendix
The function $\ln(x)$ is the inverse of the function $e^x$. More generally the function $\log_a(x)$ is the inverse of the function $a^x$ where $\log_a$ is called the logarithm to the "base" $a$. In words, what this means is if we have: $$z=x^y \tag{i}$$ then: $\log_x(z)$ is the value we would have to raise $x$ to to get $z$. We can see from $(i)$ that $y$ is the value we have to raise $x$ to get $z$ so: $$y=\log_x(z).$$ With the complicated definition out of the way, some simple examples will be more informative. If we have: $$2^a=b$$ and we want to solve for $a$ in terms of $b$ we invert the exponentiation, i.e. we invert the "raising of the base $2$ to a power", using a logarithm with the same base; in this example the base is $2$ so we apply the logarithm to the base $2$ to both sides and get get: $$a=\log_2(b).$$ Conversely, if as in the question at hand, we start with a logarithm, e.g: $$\log_{10}(a)=b$$ we can express $a$ in terms of $b$ by inverting the logarithm to the base $10$ by raising the base $10$ to the power of both sides: $$10^{\log_{10}(a)}=a=10^b.$$ The inverting of the natural (or Naperian) logarithm, $\ln$, which is log to the base $e$, was applied to both sides in the original example to "undo", and get rid of the logarithm. This was a formal step but in truth it maybe could have been skipped because it shows that if $A,B$ are positive real valued variables, i.e. they don't involve $\sqrt{-1}=i$ and are not negative because logs of negative numbers are themselves complex, then: $$\ln(A)=\ln(B)$$ implies: $$A=B.$$ Interestingly, if the variables are complex and do involve $\sqrt{-1}=i$ or are negative the "inverting" is not well defined and some further restrictions need to be specified but that is for future reading, see topics on complex logarithms.
$\endgroup$ 3 $\begingroup$For question 1. Use the property $\ln a -\ln b =\ln \frac{a}{b}$ to get $$\ln\frac{x+1}{x}=\ln3$$ from where you get $x=1/2.$
For question 2. use the property $\ln a+ \ln b= \ln ab$ to get $$\ln x(x+1)=\ln 2$$ from where you get the quadratic equation $$x^2+x=2$$ whic gives you $x=1,-2$. But since in the original equation you took log of both $x$ and $x+1$ hence, $x+1>0$, which leaves you with $x=1$.
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