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I am having trouble solving the equation

$$3e^{−x+2} = 5e^{x-1}$$

Any help would be appreciated. Thanks.

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4  
Hint: take the natural logarithm of both sides! – Nigel Overmars Jan 6 at 15:38
    
hint: Try making the problem unnecessarily difficult by turning it into a quadratic equation. – John Joy Jan 7 at 16:41
    
@JohnJoy um... I think you should either take out the part "unnecessarily difficult" or remove that comment altogether and make it an answer. – Simple Art Jan 8 at 22:10
    
@SImple Art needs to lighten up a little bit. – John Joy Jan 9 at 0:43
    
@JohnJoy LoL, ok. – Simple Art Jan 9 at 0:58

HINT:

Take log in both sides to find $$\ln3+(-x+2)=\ln5+x-1$$

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@user303134, See proofwiki.org/wiki/Laws_of_Logarithms – lab bhattacharjee Jan 7 at 5:04

$$3e^{−x+2} = 5e^{x-1}$$ Taking the natural $\log$ of both sides gives: $$\ln (3e^{−x+2}) = \ln(5e^{x-1})$$ $$\ln3+(2-x) = \ln5+(x-1) ~~~(\text{ since } \ln(ab)=\ln a + \ln b).$$ So $$\ln 3 - \ln 5+3=2x.$$ But $\ln\left(\frac{a}{b}\right)=\ln a -\ln b$. $$\therefore x=\frac{1}{2}\Big(3+\ln\left(\frac{3}{5}\right)\Big).$$

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we have $\frac{3}{5}=e^{2x-3}$ by the power rule and then $\ln(\frac{3}{5})=2x-3$ can you proceed?

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You can also make the equation much simpler just by remembering your properties of exponents: $$3e^{−x+2} = 5e^{x-1}$$ $$3e^{−x}e^{2} = 5e^{x}e^{-1}$$ multiply across by e $$3e^{-x}e^{3}=5e^{x}$$ multiply across by $e^{x}$ and divide by $5$, take the natural log of both sides... $$3e^{3} = 5e^{2x}$$ Now you've got the $x$ on one side, :) rest ain't so bad $$\frac{3}{5}e^{3} = e^{2x}$$ $$\ln(\frac{3}{5}e^{3}) = \ln(e^{2x})$$ $$\ln(\frac{3}{5}e^{3}) = 2x$$ $$\therefore x = \frac{1}{2}\ln(\frac{3}{5}e^{3})$$

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$$3e^{-x+2} = 5e^{x-1}$$ $$\frac{3e^2}{e^x} = \frac{5e^x}{e^1}$$ $$\dots = \dots$$

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