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While trying to answer this SO question I got stuck on a messy bit of algebra: given

$$ \log m = \log n + \frac32 \, \log \biggl( 1 + \frac{v}{m^2} \biggr) $$

I need to solve for $m$. I no longer remember enough logarithmic identities to attempt to do this by hand. Maxima can’t do it at all, and Wolfram Alpha coughs up a hairball that appears to be the zeroes of a quartic, with no obvious relationship to the original equation.

Is there a short, tidy solution? Failing that, an explanation of how WA managed to turn this into a quartic, and the quartic itself, would be ok.

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  • $\begingroup$ Hint: Put everything on one side of the equation and combine the logarithms using $\log a + \log b= \log(a b)$. $\endgroup$
    – Fabian
    Jan 4 '17 at 14:59
  • $\begingroup$ It's a quartic equation. If you have changed your equation into the form $a(m^2)^4+b(m^2)^3+c(m^2)^2+d(m^2)+e=0$ you can solve explicitly, see e.g. en.wikipedia.org/wiki/Quartic_function . Otherwise use an iteration. $\endgroup$
    – user90369
    Jan 4 '17 at 15:31
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$$\log {m}=\log {n}+\frac{3}{2} \log {\left(1+\frac{v}{m^2}\right)}$$ $$\log {m}=\log {n}+ \log {\left(\left(1+\frac{v}{m^2}\right)^{\frac{3}{2}}\right)}$$ $$\log {m}=\log {\left({n\cdot\left(1+\frac{v}{m^2}\right)^{\frac{3}{2}}}\right)}$$ Exponentiate both sides: $$m=n\cdot\left(1+\frac{v}{m^2}\right)^{\frac{3}{2}}$$ $$m^2=n^2\cdot\left(1+\frac{v}{m^2}\right)^3$$ $$m^2=n^2\cdot\left(1+\frac{3v}{m^2}+\frac{3v^2}{m^4}+\frac{v^3}{m^6}\right)$$ $$m^2=\frac{n^2}{m^6}\left(m^6+3v\cdot m^4+3v^2 \cdot m^2+v^3\right)$$ $$m^8=n^2 m^6+3vn^2 m^4+ 3n^2v^2 m^2+n^2v^3$$ Substitute $m^2=u$ and you will obtain a quartic expression. $$u^4-n^2u^3-3vn^2 u^2-3n^2 v^2 u-n^2v^3=0$$ This is going to be a long solution, however you can use the general formula for the solution to a quartic equation.

Wikipedia suggests using substitutions in order to solve it.

The full formula without substitution for $ax^4+bx^3+cx^2+dx+e$ is below (I cannot write it in $\LaTeX$ because it is so long).

General formula without substitution

This suggests that your solution in Wolfram Alpha is probably correct.

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If by solve you mean isolate $m$ in terms of $n$ and $v$ you can use the one-to-one property of logarithms: \begin{align*} \log m & = \log n + \frac{3}{2} \log \left(1 + \frac{v}{m^2}\right) \\ \log\left(\frac{m}{n}\right) & = \log\left(\left(1+\frac{v}{m^2}\right)^\frac{3}{2}\right) \\ \frac{m}{n} & = \left(1+\frac{v}{m^2}\right)^\frac{3}{2} \\ \frac{m^2}{n^2} & = \left(1+\frac{v}{m^2}\right)^3 \end{align*}

Now the problem is simple algebra in solving a cubic in $m^2$.

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$$\log m = \log n + \frac32 \, \log \biggl( 1 + \frac{v}{m^2} \biggr)=\log n+\log \biggl( 1 + \frac{v}{m^2}\biggr)^{3/2}=\log n\biggl( 1 + \frac{v}{m^2}\biggr)^{3/2}$$

Then

$$m=n\biggl( 1 + \frac{v}{m^2}\biggr)^{3/2} \rightarrow m^{2}=\frac{n^2(m^2+v)^3}{m^6}\rightarrow m^8=n^2(m^2+v)^3$$

I just don't know if the above equation is solvable for $m$.

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  • $\begingroup$ I think it's $m^{2}=\frac{n^2(m^2+v)^3}{m^6}$. $\endgroup$
    – Blencer
    Jan 4 '17 at 15:10
  • $\begingroup$ @ArnaldoNascimento I think: $$\text{m}^2=\left(\text{n}\cdot\left(1+\frac{\text{v}}{\text{m}^2}\right)\right)^2=\text{n}^2\cdot\frac{\left(\text{m}^2+\text{v}\right)^2}{\text{m}^4}\space\Longleftrightarrow\space\text{m}^2\cdot\text{m}^4=\text{m}^6=\text{n}^2\cdot\left(\text{m}^2+\text{v}\right)^2$$ $\endgroup$ Jan 4 '17 at 15:34
  • $\begingroup$ Hi @JanEerland, your first equality is wrong. Please, check again. $\endgroup$
    – Arnaldo
    Jan 4 '17 at 15:44
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$logm$=$logn$+$3/2log(1+v/m^2)$
$logm$=$logn$+$log((m^2+v)/m^2)$
$logm$=$logn$+$log(m^2+v)^(1.5))$
$logm$=$logn$+$log(m^2+v)^3/2$-$logm^3$
$logm^3$+$logm$-$logn$=$log(m^2+v)^3/2$
$log(m^4/n)$=$log(m^2+v)$
$m^4/n$=$(m^2+v)^(3/2)$ From here think you gotta use the cubic expension and square out the root It is my first answer so plz dont judge me...

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  • $\begingroup$ can anyone teach me how to type 3/2 on square root? it becomes like cubic then divide by two $\endgroup$
    – M. Chen
    Jan 4 '17 at 15:18
  • $\begingroup$ Use "(m^2+v)^{3/2}" or "(m^2+v)^{\frac{3}{2}}". (Using these parantheses "{}") $\endgroup$ Jan 4 '17 at 15:35
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Assuming that all variables are real and positive:

$$\ln\text{m}=\ln\text{n}+\frac{3}{2}\cdot\ln\left(1+\frac{\text{v}}{\text{m}^2}\right)\space\Longleftrightarrow\space\frac{2}{3}\cdot\left(\ln\text{m}-\ln\text{n}\right)=\ln\left(1+\frac{\text{v}}{\text{m}^2}\right)$$

Now, use:

$$\ln\text{a}-\ln\text{b}=\ln\frac{\text{a}}{\text{b}}$$

So, we get:

$$\frac{2}{3}\cdot\ln\frac{\text{m}}{\text{n}}=\ln\left(1+\frac{\text{v}}{\text{m}^2}\right)$$

Now, use:

$$\exp\left(\text{c}\cdot\ln\text{d}\right)=e^{\text{c}\cdot\ln\text{d}}=\text{d}^\text{c}$$

So, we get (taking the $\exp$ of both sides):

$$\exp\left\{\frac{2}{3}\cdot\ln\frac{\text{m}}{\text{n}}\right\}=\exp\left\{\ln\left(1+\frac{\text{v}}{\text{m}^2}\right)\right\}=\color{red}{\left(\frac{\text{m}}{\text{n}}\right)^{\frac{2}{3}}=1+\frac{\text{v}}{\text{m}^2}}$$

Now, we want to solve $\text{m}$:

$$\left(\frac{\text{m}}{\text{n}}\right)^{\frac{2}{3}}=\frac{\text{m}^{\frac{2}{3}}}{\text{n}^{\frac{2}{3}}}=1+\frac{\text{v}}{\text{m}^2}=\frac{\text{m}^2+\text{v}}{\text{m}^2}\space\Longleftrightarrow\space\text{m}^8=\text{n}^2\cdot\left(\text{m}^2+\text{v}\right)^3$$

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