Can someone solve this for me please? Use natural log to find x...?

Question

Could someone please solve the following equation for x, showing all steps?

0 = 3(e^3x) - 2(e^-2x)

Note: The final answer should be (1/5)(ln(2/3)), but I'm not sure how you get there.

Answer 1

Your first step is to convert all negative exponents into positive exponents. Keep in mind that this can be accomplished by putting one over the term with the exponent.

0 = 3(e^[3x]) - 2/(e^[2x])

Now, merge them into one fraction

0 = { 3(e^[3x])(e^[2x]) - 2 } / e^[2x]

Remember that when we multiply two numbers where the bases are the same but the exponents differ, we can add the exponents.

0 = { 3(e^[3x + 2x]) - 2 } / e^[2x]

And nowe we simplify ...

0 = { 3e^[5x] - 2 } / e^[2x]

Now, we can multiply both sides of the equation by e^(2x). This is an entirely safe thing to do, because e^(2x) is always going to be a positive number and will never be 0 (as the graph e^[2x] has a horizontal asymptote of 0).

0 = 3e^(5x) - 2

Now, bring the 2 to the left hand side,

2 = 3e^(5x)

Divide both sides by 3,

2/3 = e^(5x)

And now, convert to logarithmic form. When doing this, remember that the key thing to remember upon conversion is that the base of the log becomes the base of the exponent. Every thing else you should memorize (like where the 2/3 goes and where the 5x goes subsequent to the form transformation).

log [ base e ] (2/3) = 5x

Remember that "log base e" is the natural log, or ln

ln(2/3) = 5x

And now we divide both sides by 5.

ln(2/3) / 5 = x

or

x = (1/5) ln (2/3)

Answer 2

3 e^x - 2/ e^x = 4 Taking LCM 3 e^2x - 2 = 4 e^x Rearranging 3 e^2x - 4 e^x - 2 = 0 This is a quadriatic equation Put e^x = p 3 p^2 - 4p - 2 = 0 Solving e^x = (2 + sqrt 10)/ 3 and ( 2 - sqrt 10)/3 Taking log on both sides x log e = Log ( 2 + sqrt 10) - log 3 To convert to log base 10 ignore log e as we have to multiply each term with 2.303 Rest of the answer is an arithmetic work. pleas do it and find.

Answer 3

here are the steps:

1) 0=3(e^3x) - 2(e^-2x)

2) 2(e^-2x) = 3(e^3x)

3) 2 (1/e^2x) = 3(e^3x)

4) 2/3= e^3x * e^2x

5) 2/3 = e^(3+2)x

6) 2/3 = e^5x

taking log base e on both sides it is reduced to

5x= ln(2/3)

7) finally x= (1/5)(ln(2/3)

njoi

Answer 4

First rearrange the terms:

3(e^3x) = 2(e^-2x)

Take the natural log of both sides, remembering that the log of a product is the sum of the logs, so you get

ln(3) + ln(e^3x) = ln(2) + ln(e^-2x)

The ln of e to a power is the value of that power: ln(e^a) = a. so

ln(3) + 3x = ln(2) = -2x;

solve for x: 5x = ln(2) - ln(3)
x = [ln(2) - ln(3)]/5 = ln(2/3)/5

Answer 5

Rearranging,
    3(e^3x) = 2(e^-2x)
multiply both sides by (e^2x),
    3(e^5x) = 2(1)
divide both sides by 3,
    e^5x = 2/3
natural log,
    ln(e^5x) = ln(2/3)
from which,
    5x = ln(2/3)
or,
    x = (1/5)ln(2/3)

Answer 6

0 = 3(e^3x) - 2(e^-2x)

To solve for x, rearrange the terms thusly:

2(e^-2x) = 3(e^3x)

Multiply both sides by e^2x

2 = 3(e^5x)
ln 2 = ln 3(e^5x) = ln 3 + ln e^5x
ln 2 - ln 3 = ln e^5x = 5x ln e = 5x
(1/5)(ln 2/3) = x

x = (1/5)(ln 2/3) = -0.081093

Answer 7

0 = 3(e^3x) - 2(e^-2x)
3(e^3x) = 2(e^-2x)
2/3 = (e^3x)/(e^-2x)
2/3 = e^(3x-(-2x))
2/3 = e^5x
ln(2/3) = lne^5x
ln(2/3) = 5xlne
ln(2/3) = 5x
x = (1/5)(ln(2/3))

Answer 8

dude go to ask .com and type that in and put something like how do i find the steps for or how do i find the answer for...it should work

www.ergra.com/Algebra-help-20-6.htm

try that one man hope i helped

Answer 9

3(e^3x)-2(e^-2x)=0
3(e^3x)=2(e^-2x)=0
3x+ln3=-2x+ln2
3x+2x=ln2-ln3
x(3+2)=ln(2/3)
5x=ln(2/3)
x=ln(2/3)/5