Solving Exponential Equations: e^2x = 5e^3x

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e^2x = 5e^3x


I understand that I need to take a natural log of both sides here, what I am thrown about is the constant "5". Can I bring that up as an exponent? So, e^2x = e^(3x)^5?
 
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No, not really. That's a rule for logarithms, not for exponents.

However, it's not really a problem. Once you take the natural log of both sides, you just need to remember a different logarithm rule:

[tex]log(ab)=log(a)+log(b)[/tex]

That's all you need here.
 
You might also rewrite this equation, dividing with 5e^2x on both sides:
e^x=1/5
 
Sorry but I don't completely follow. I get that I need to take a natural log of both sides, so are you saying: ln(e^2x) = 5 ln(e^2x)?
 
cue928 said:
Sorry but I don't completely follow. I get that I need to take a natural log of both sides, so are you saying: ln(e^2x) = 5 ln(e^2x)?

No, I'm saying that ln(e^2x)=ln(5e^(3x)). From there, you can use the logarithm rule I posted above to separate the 5.
 
Char. Limit said:
No, I'm saying that ln(e^2x)=ln(5e^(3x)). From there, you can use the logarithm rule I posted above to separate the 5.

Thanks, got it!