How to Solve Exponential Equations with Natural Logs?

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To solve the exponential equation 50e^(2x) = 100e^(0.5x), start by dividing both sides by 50, resulting in e^(2x) = 2e^(0.5x). Taking the natural logarithm of both sides leads to the equation 2x = ln(2) + 0.5x. Rearranging gives 1.5x = ln(2), which simplifies to x = ln(2)/1.5. This method effectively uses natural logs to isolate x in the exponential equation.
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hey
2day i had a test in intro calc and was given the following problem to solve:

50e^(2x) = 100e^(0.5x)

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:

50e^2.5x = 100e^x

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,

50e^{2x} = 100e^{0.5x}

start by dividing the equation by 50

e^{2x} = 2e^{0.5x}

take natural log

2x = ln 2\,+\,0.5x

1.5x = ln 2

x = \frac{ln 2}{1.5}
 
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Astronuc said:
danago is correct.
I just didn't want to give the game away :smile:
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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