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### AuthorTopic: How do I solve for x?  (Read 303 times) Tweet Share

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#### Jesse_551

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##### How do I solve for x?
« on: October 12, 2019, 06:25:04 pm »
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This is the question:

Solve 2e^-x + 5 = 3e^-x for x.

It would be much appreciated, thanks.

#### DrDusk

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##### Re: How do I solve for x?
« Reply #1 on: October 12, 2019, 06:42:05 pm »
+2
This is the question:

Solve 2e^-x + 5 = 3e^-x for x.

It would be much appreciated, thanks.

$2e^{-x} + 5 = 3e^{-x} \\ \Rightarrow 3e^{-x} -2e^{-x} = 5 \\ \Rightarrow e^{-x}(3-2) = 5 \\ \Rightarrow ln(e^{-x}) = ln(5) \\ \Rightarrow -x = ln(5) \\ \therefore x = -ln(5)$
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#### Jesse_551

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##### Re: How do I solve for x?
« Reply #2 on: October 13, 2019, 09:01:28 am »
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$2e^{-x} + 5 = 3e^{-x} \\ \Rightarrow 3e^{-x} -2e^{-x} = 5 \\ \Rightarrow e^{-x}(3-2) = 5 \\ \Rightarrow ln(e^{-x}) = ln(5) \\ \Rightarrow -x = ln(5) \\ \therefore x = -ln(5)$

Thank you, can you just clarify steps 2 to 3 I don’t understand how you got from:

e^-x(3-2)=5   to  ln(e^-x)=ln(5)

Thanks again

#### AngelWings

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##### Re: How do I solve for x?
« Reply #3 on: October 13, 2019, 01:17:14 pm »
+2

Thank you, can you just clarify steps 2 to 3 I don’t understand how you got from:

e^-x(3-2)=5   to  ln(e^-x)=ln(5)

Thanks again
I’m not DrDusk, but I think they did 3 - 2 in the brackets and then used the opposite operation of e, i.e. loge or ln in one step.
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#### DrDusk

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##### Re: How do I solve for x?
« Reply #4 on: October 13, 2019, 01:48:41 pm »
0

Thank you, can you just clarify steps 2 to 3 I don’t understand how you got from:

e^-x(3-2)=5   to  ln(e^-x)=ln(5)

Thanks again
Yep AngleWings said it.
Because 3-2 = 1 so that because 1*e^-x = 5. Now you take ln which is log base e of both sides which is something you can do and should do as it simplifies any exponential of the form e^(Something).
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