Pre Calculus problem in need of help

[TheKracken]

Homework Statement

Solve for X in terms of Natural Logs;
Both my friend and I (top scores in the class) are struggling with this problem.

Homework Equations

e^ax=c*2^bx

The Attempt at a Solution

We both decided to divide both sides by 2^bx which gave us
e^ax / 2^bx = C

Then I flipped the 2^bx up to the top by making the ^-bx
Then I multiplied the 2 and the e and added the ^ax and ^-bx

thus resulting in
2e^ax-bx=c

Then divide both sides by two and natural log both sides
getting

ax-bx= ln(c/2)

Factor out a x and divide both sides by a-b

x= ln(c/2) / a-b

We both suspect our answers are wrong due to the multiplying of 2 and the e, other than that we have no idea how to get the right answer.

 

[SammyS]

Homework Statement

Solve for X in terms of Natural Logs;
Both my friend and I (top scores in the class) are struggling with this problem.

Homework Equations

e^ax=c*2^bx

The Attempt at a Solution

We both decided to divide both sides by 2^bx which gave us
e^ax / 2^bx = C

Then I flipped the 2^bx up to the top by making the ^-bx
Then I multiplied the 2 and the e and added the ^ax and ^-bx

thus resulting in
2e^ax-bx=c

Then divide both sides by two and natural log both sides
getting

ax-bx= ln(c/2)

Factor out a x and divide both sides by a-b

x= ln(c/2) / a-b

We both suspect our answers are wrong due to the multiplying of 2 and the e, other than that we have no idea how to get the right answer.

You're right regarding the error of multiplying 2 and e .

The general way to solve an exponential equation is to take some logarithm of both sides. In this case using the natural log, ln, should work fine.

 

[TheKracken]

Trying what I believe you just said still confuses me and I am still not getting an answer. Would you possibly explain further?

 

[SammyS]

Trying what I believe you just said still confuses me and I am still not getting an answer. Would you possibly explain further?

Show what you get when you take ln of both sides of

e^ax=c*2^bx​

(Assuming that's the equation you started with. You really should have used parentheses to set off the exponents.)

 

[TheKracken]

Alright, so my results are now looking like this...

x= (ln(2c))/(a+(1/b))

Is this now correct?
(I do not have an answer key for this)

 

[SammyS]

Alright, so my results are now looking like this...

x= (ln(2c))/(a+(1/b))

Is this now correct?
(I do not have an answer key for this)

No.

Show some of your steps.

2 and c shouldn't combine like that at all.

 

[TheKracken]

I took the ln of both sides, then the ln(e^(ax)) went to ax=ln(c*2^(bx)) From here I put the bx in front of the ln(c*2) but now I suspect that is wrong as well. I am not sure what to do with the ln(c*2^(bx))

 

[micromass]

I took the ln of both sides, then the ln(e^(ax)) went to ax=ln(c*2^(bx)) From here I put the bx in front of the ln(c*2) but now I suspect that is wrong as well. I am not sure what to do with the ln(c*2^(bx))

\log(ab) = \log(a) + \log(b)

 

[TheKracken]

Right, I feel pretty stupid for that now.

Ok well, now...
ax= ln(c) + ln(2^(bx))
Bring the bx to the front and subtract to the other side.
ax-bxln(2)=ln(c)

Factor out x

x(a-bln(2))= ln(c)

divide by (a-bln(2))
x= ln(c)/(a-bln(2))

Alright...now I feel like I am getting closer to the answer, if not finally hit it.

 

[Mark44]

Right, I feel pretty stupid for that now.

Ok well, now...
ax= ln(c) + ln(2^(bx))
Bring the bx to the front and subtract to the other side.
ax-bxln(2)=ln(c)

Factor out x

x(a-bln(2))= ln(c)

divide by (a-bln(2))
x= ln(c)/(a-bln(2))

That's it.

Alright...now I feel like I am getting closer to the answer, if not finally hit it.