how do you solve this equation : xe^(3x) - 1 = 5 ? algebraically.?

Answer 1

(3x)xe1=
3x3x2 e^=5+1=6/3
x2 e1=6/3=2
x2=2 e1
or
^(3x) = 6/x ( x != 0)

3x = ln (6/x) = ln 6 - ln x

3x + ln x = ln 6

x = (ln 6 - ln x) / 3

I don't think you can solve it any further algebraically.
You could solve it numerically.
Take x = 1 (or any other number) and substitute it in the right part.

You get:
x = 0.5973
Fill this in again (and repeat this):
x = 0.7691
x = 0.6848
x = 0.7235
...
x = 0.7110 -> this is a stable point

x = 0.7110 is the solution (filling this in in the original equation verifies this)

Answer 2

By Addition property of Equality:
xe^(3x) - 1 = 5 ==> xe^(3x) = 6
Dividing both side of the equation by x we have:
xe^(3x) = 6 ==> e^(3x) = 6/x
Applying the laws of logarithms:
e^(3x) = 6/x ==> 3x = ln (6/x)
==> 3x = ln 6 - ln x
==> ln x + 3x = ln 6
Algebra approach stops here we cant solve further since the x's can't be group together. But we can solve this by graphical approach or analytical approach (using calculus, i.e., by newton method)
Using the either of the two approach you will arrive at this x = 0.7110 approximate up to 4 decimal places.

Answer 3

THere are 2 procedures to unravel. Rewrite y in words of x or make certain concurrently. Doing the latter you get 3[5x 5x + 2y = a million] 2[4x + 3y = 5 = a million] 2[4x + 3y = 5] 15x + 6y = 3 8x + 6y = 10 Subtract and also you get 7x = -7 x = -a million change -a million in for x in both equation 4(-a million) + 3y = 5 -4 + 3y = 5 3y = 9 y = 3

Answer 4

The simple fact of the matter is that you can't. It's not possible to group the x terms together and isolate the x. Just like it is impossible for these other equations:

x + ln(x) = 6
x^2 + sin(x) = 2

The only way you can solve this is numerically.

Had it been equal to 0, i.e.
xe^(3x) = 0, then we can just equate each factor to 0 and get solutions. But that's definitely not the case here.

Answer 5

e^(3x) = 6/x ( x != 0)

3x = ln (6/x) = ln 6 - ln x

3x + ln x = ln 6

x = (ln 6 - ln x) / 3

I don't think you can solve it any further algebraically.
You could solve it numerically.
Take x = 1 (or any other number) and substitute it in the right part.

You get:
x = 0.5973
Fill this in again (and repeat this):
x = 0.7691
x = 0.6848
x = 0.7235
...
x = 0.7110 -> this is a stable point

x = 0.7110 is the solution (filling this in in the original equation verifies this)

update:
this is indeed W(18)/3 as one of the previous answerers states but you can only evaluate this function iteratively

Answer 6

The exponential function is known as a trancendental function. The definition of a transcendental function is one which cannot be solved algebraically.

Not possible to solve algebraically.

Answer 7

x = 6e^-3x
iterating,
est. calc % error
0.50000 1.33878 0
0.91939 0.38045 71.58268
0.64992 0.85385 -124.43529
0.75189 0.62883 26.35429
0.69036 0.75631 -20.27237
0.72333 0.68507 9.41874
0.7042 0.72554 -5.9069
0.71487 0.70268 3.14978
0.70878 0.71565 -1.8445
0.71221 0.70831 1.02502
0.71026 0.71247 -0.58689
0.71136 0.71011 0.33041
0.71074 0.71145 -0.18781
0.71109 0.71069 0.10618
0.71089 0.71112 -0.06021
0.71101 0.71088 0.03408
0.71094 0.71101 -0.01931

x = 0.711

Answer 8

the previous answer (sri_july27)has found the maximum or minimum value of the function f(x)= xe^3x.
no equation should be slved by differentiating......for eg consider x^2 = 4.....if u diff ull get 2x=0.....x=0....thats not the solution.....i know it doesnt answer ur question....but i jus wanted to say that the prev answer is wrong....

Answer 9

You can't do it using just the functions you probably learned in school. However, if you know the lambert W function, then it's easy:

xe^(3x) - 1 = 5
xe^(3x) = 6
3xe^(3x) = 18
3x=W(18)
x=W(18)/3

Answer 10

probably try this:

xe^(3x)=6

differentiate the equation with respect to x

so u get

e^3x+x. 3 .e^ 3x= 0

e^3x ( 1+3x)=0

e^3x cannot be zero , cuz if it is zero the firat equoation will become zero which is not possible.

so
1+3x=0
which means

x= -1/3