# Solve the equations for x.

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

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

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

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

$$In(\frac{7}{10})+axIn(2)=2x$$

$$In(\frac{7}{10})=2x-axIn(2)$$

$$In(\frac{7}{10})=x(2-aIn(2))$$

$$x=\frac{In(\frac{7}{10})}{2-aIn(2)}$$

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

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

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

$$2(a-x)=3x$$

$$2a-2x=3x$$

$$2a=5x$$

$$x=\frac{2a}{5}$$

• That is Correct – user35687 Aug 14 '16 at 18:07
• Thanks. Do you know any better way or faster? – 关一骏 Aug 14 '16 at 18:08
• I think you found the best way to solve them, congratulations! – Peter Aug 14 '16 at 18:09
• In the answer below, it is pointed out that we have to be careful in exercise $a)$ (denominator could be $0$) – 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 $\text{a}\ne0$.

• You are right with $a)$ . In $b)$ , the case $a=0$ is valid. – Peter Aug 14 '16 at 18:14
• @Peter That was what I was pointing out here. – Jan Eerland Aug 14 '16 at 18:14
• 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: $$2\cdot\frac{0-x}{3\cdot\sqrt[3]{0}}-0^{\frac{2}{3}}$$ – Jan Eerland Aug 14 '16 at 18:20
• @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 – Jan Eerland Aug 14 '16 at 18:22
• @C.Guan You're welcome! – Jan Eerland Aug 14 '16 at 18:29