How to Solve Exponential Equations with Natural Logs?

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pavadrin
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hey
2day i had a test in intro calc and was given the following problem to solve:

[tex]50e^(2x) = 100e^(0.5x)[/tex]

i was quite annoyed :mad: at this problem because i didn’t know how to solve it. although i do know that natural logs need to be taken on both sides to solve it, however the coefficient of the e really had me confused. any help would be greatly appreciated, thanks,
Pavadrin
 
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What happens if you multiply both sides by exp(-0.5x)... ?
 
I got stuck on the same question :(

heres the working i did for it after the test.

http://img145.imageshack.us/img145/3503/logszp0.gif
 
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if both sides are multiplied by exp(-0.5x) would that make it:

[tex]50e^2.5x = 100e^x[/tex]

but i don't see how that helps, thanks for the relpy though
 
eax * e-ax = eax-ax = e0 = 1

danago is correct.

Using ln,

[tex]50e^{2x} = 100e^{0.5x}[/tex]

start by dividing the equation by 50

[tex]e^{2x} = 2e^{0.5x}[/tex]

take natural log

[tex]2x = ln 2\,+\,0.5x[/tex]

[tex]1.5x = ln 2[/tex]

[tex]x = \frac{ln 2}{1.5}[/tex]
 
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Astronuc said:
danago is correct.
I just didn't want to give the game away :smile: