Equations-EMT

Algebra Level 2

$\large \begin{cases} e^{mt} = emt+mt^e-1 \\ e^{mt} = emt+tm^e-1 \\ \ln(emt)=1 \end{cases}$

Clarification : $e \approx 2.71828$ denotes the Euler's number .

Let $m$ and $t$ be positive real numbers satisfying the system of equations above, find $m+t$ .

1 solution

Tommy Li
Dec 22, 2016

$\large \begin{cases} e^{mt} = emt+mt^e-1 \\ e^{mt} = emt+tm^e-1 \\ \ln(emt)=1 \end{cases}$

$\large \begin{cases} mt^e=tm^e \\ emt= e \end{cases}$

$\large \begin{cases} \ln(m) +e\ln(t) = \ln(t) +e\ln(m) \\ mt= 1 \end{cases}$

$\large \begin{cases} e\ln(t) -e\ln(m) = \ln(t) -\ln(m) \\ mt= 1 \end{cases}$

$\large \begin{cases} (e-1)(\ln(t) -\ln(m)) =0 \\ mt= 1 \end{cases}$

$\large \begin{cases} \ln(t) =\ln(m) \\ mt= 1 \end{cases}$

$\large \begin{cases} m=t \\ t^2 = 1 \end{cases}$

$\large \begin{cases} m = 1 \\ t = 1 \end{cases}$ or $\large \begin{cases} m = -1 \\ t = -1 \end{cases}$ (rej.)

$\Rightarrow m + t = 1+1 = 2$

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Tommy Li - 4 years, 5 months ago