a. $10e^{2x}=7\cdot 2^{ax}$

b. $\frac{2(a-x)}{3\sqrt[3]{x}}-x^{2/3}=0$

I did both, but I'm not sure if I did them correctly. Here is my work for both questions:

a. $10e^{2x}=7\cdot 2^{ax}$

$e^{2x}=\frac{7\cdot 2^{ax}}{10}$

$In(\frac{7\cdot 2^{ax}}{10})=2x$






b. $\frac{2(a-x)}{3\sqrt[3]{x}}-x^{2/3}=0$


$2(a-x)=3x^{1/3}\cdot x^{2/3}$





  • 2
    $\begingroup$ That is Correct $\endgroup$ – user35687 Aug 14 '16 at 18:07
  • $\begingroup$ Thanks. Do you know any better way or faster? $\endgroup$ – 关一骏 Aug 14 '16 at 18:08
  • 2
    $\begingroup$ I think you found the best way to solve them, congratulations! $\endgroup$ – Peter Aug 14 '16 at 18:09
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    $\begingroup$ In the answer below, it is pointed out that we have to be careful in exercise $a)$ (denominator could be $0$) $\endgroup$ – Peter Aug 14 '16 at 18:15

For (a), the real solution (correct):

$$10e^{2x}=7\cdot2^{\text{a}x}\Longleftrightarrow$$ $$\ln(10e^{2x})=\ln(7\cdot2^{\text{a}x})\Longleftrightarrow$$ $$\ln(10)+\ln(e^{2x})=\ln(7)+\ln(2^{\text{a}x})\Longleftrightarrow$$ $$2x+\ln(10)=\ln(7)+\text{a}x\ln(2)\Longleftrightarrow$$ $$2x-\text{a}x\ln(2)=\ln(7)-\ln(10)\Longleftrightarrow$$ $$x(2-\text{a}\ln(2))=\ln(7)-\ln(10)\Longleftrightarrow$$ $$x=\frac{\ln(7)-\ln(10)}{2-\text{a}\ln(2)}$$

Where $\text{a}\ln(2)\ne2$

(b) is also correct, but $\text{a}\ne0$.

  • $\begingroup$ You are right with $a)$ . In $b)$ , the case $a=0$ is valid. $\endgroup$ – Peter Aug 14 '16 at 18:14
  • $\begingroup$ @Peter That was what I was pointing out here. $\endgroup$ – Jan Eerland Aug 14 '16 at 18:14
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    $\begingroup$ Now, look when $a=0$: $$2\cdot\frac{0-x}{3\cdot\sqrt[3]{x}}-x^{\frac{2}{3}}=0\Longleftrightarrow-\frac{5x^{\frac{2}{3}}}{3}=0$$ That gives us a solution at $x=0$, but when we check that: $$2\cdot\frac{0-x}{3\cdot\sqrt[3]{0}}-0^{\frac{2}{3}}$$ $\endgroup$ – Jan Eerland Aug 14 '16 at 18:20
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    $\begingroup$ @Peter You're welcome, when we use a limit then we can get to zero but it has to be in a carefull way! Otherwise you allowed yourself to divide by zero .... kuch $\endgroup$ – Jan Eerland Aug 14 '16 at 18:22
  • 1
    $\begingroup$ @C.Guan You're welcome! $\endgroup$ – Jan Eerland Aug 14 '16 at 18:29

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