Log equation with two raised variables

[QuarkCharmer]

1. The problem statement, all variables and given/known data
This is not a specific problem that I must complete, but I realized that I forgot how to solve these style problems!

2. Relevant equations

3. The attempt at a solution

For instance, off the top of my head:
3e^{x+1}=e^{x+2}

I would divide both sides through by 5 to get:
3e^{x+1}=e^{x+2}
^This should look like: 3e^(x+1)=e^(x+2), I don't know what is up with latex.

Then I would probably take the natural log of both sides, but that coefficient is messing up my idea. Can I do this?
(x+1)ln3e = (x+2)lne
(x+1)ln3e = (x+2)
xln3e + ln3e = x+2
x(ln3e + ln3e) = x+2
x(ln3e + ln3e) - x = 2
x((ln3e + ln3e) - 1 = 2
ln3e + ln3e - 1 = \frac{2}{x}

If that is even possible, which I highly doubt, I am stuck.

[Pranav-Arora]

Is the question this:-

15e^{x+1} = 5e^{x+2}

[QuarkCharmer]

Yes it is.

[Mark44]

1. The problem statement, all variables and given/known data
This is not a specific problem that I must complete, but I realized that I forgot how to solve these style problems!

2. Relevant equations

3. The attempt at a solution

For instance, off the top of my head:
15e^{x+1}=5e^{x+2}
Fixed the LaTeX in the equation above. Here's what you do.
15e^{x+1} - 5e^{x+2} = 0
15e\cdot e^x - 5e^2 \cdot e^x = 0
5e^x(3e - e^2 ) = 0
Then either 5ex = 0 or 3e - e2 = 0, neither of which can happen.




I would divide both sides through by 5 to get:
3e^(x+1)=e^(x+2)
^This should look like: 3e^(x+1)=e^(x+2), I don't know what is up with latex.

Then I would probably take the natural log of both sides, but that coefficient is messing up my idea. Can I do this?
(x+1)ln3e = (x+2)lne
(x+1)ln3e = (x+2)
xln3e + ln3e = x+2
x(ln3e + ln3e) = x+2
x(ln3e + ln3e) - x = 2
x((ln3e + ln3e) - 1 = 2
ln3e + ln3e - 1 = \frac{2}{x}

If that is even possible, which I highly doubt, I am stuck.

[QuarkCharmer]

I repaired the latex, thank you, the code seems to have changed on me!

I don't really mean (this) specific problem, just anything with the general form
e^x+a = e^x+b

Always unsolvable?

Edit: Now that I think about it, I think one of the exponents was negative in the problems I am talking about.
e^x = e^-x

[Mark44]

I repaired the latex, thank you, the code seems to have changed on me!

I don't really mean (this) specific problem, just anything with the general form
e^x+a = e^x+b
If you write them that way, you need parentheses, as in
e^(x + a) = e^(x + b)

At least that's what I think you mean.
With this equation, you can take the log of both sides, to get x + a = x + b, which is a true statement iff a = b.


Always unsolvable?

Edit: Now that I think about it, I think one of the exponents was negative in the problems I am talking about.
e^x = e^-x

Again, take the log of each side, which results in the equation x = -x, or 2x = 0, or x = 0.

[QuarkCharmer]

x=-x and 2x=0 isn't really possible unless x = 0, but I don't understand how you methodically arrived at that conclusion. How would you solve these types algebraically if possible, and how would you solve a similar problem with the coefficient attached such that:
2e^{(x+1)} = e^{-(x+2)}

[Pranav-Arora]

x=-x and 2x=0 isn't really possible unless x = 0, but I don't understand how you methodically arrived at that conclusion. How would you solve these types algebraically if possible, and how would you solve a similar problem with the coefficient attached such that:
2e^{(x+1)} = e^{-(x+2)}

I tried to solve it like this:-

2e^{(x+1)} = e^{-(x+2)}

\Rightarrow2=\frac{e^{-x-2}}{e^{x+1}}

\Rightarrow2=e^{-2x-3}

\Rightarrow\log_e 2=-2x-3

Try solving it further :smile:

[QuarkCharmer]

I see how to solve that, but I am unsure how you reached this step?
2=e^{-2x-3}

How does
\frac{e^{-x-2}}{e^{x+1}} = e^{-2x-3}

You are just subtracting the exponents? I guess I never considered that lol.

[Pranav-Arora]

I see how to solve that, but I am unsure how you reached this step?
2=e^{-2x-3}

How does
\frac{e^{-x-2}}{e^{x+1}} = e^{-2x-3}

You are just subtracting the exponents? I guess I never considered that lol.

Yeah, i am subtracting the exponents....:smile:

[Mark44]

I see how to solve that, but I am unsure how you reached this step?
2=e^{-2x-3}

How does
\frac{e^{-x-2}}{e^{x+1}} = e^{-2x-3}

You are just subtracting the exponents? I guess I never considered that lol.
If you are brushing up on solving exponential equations, you really ought to review the basic rules of exponents and the rules of logarithms...

[QuarkCharmer]

If you are brushing up on solving exponential equations, you really ought to review the basic rules of exponents and the rules of logarithms...

I came to the same conclusion.