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math brainlet needs help

Anonymous Fri 02 Oct 20:37:15 2020 No.12187003 View Reply Original Report

Quoted By: >>12187007 >>12189094 >>12191330

whenever I try to solve an equation I seem to find the most complicated way to do so. Sometimes I´m able to solve the equation but I get the wrong result. I can´t seem to find the most direct way or do the calculations in the right order.
How can I learn to find the best, most direct and correct way when solving equations?
Do you have any tips?

Anonymous

Anonymous Fri 02 Oct 2020 20:38:43 No.12187007 Report

Quoted By: >>12187136

>>12187003
could you give us some example equations to demonstrate

Anonymous

Anonymous Fri 02 Oct 2020 20:58:03 No.12187089 Report

Quoted By: >>12189207

My version:
ln(3x)-ln(1,5)=2 | e^
3x-1,5 = e^2 | +1,5
3x = e^2 + 1,5 | /3
x = e2+ 1,5 / 3

I should have done
ln(3x/1,5) = 2 instead of using e in the beginning.

My Version:
ln(3^x)= ln (27)
x*ln(3) = ln (27) | e^
e^x*3 = 27 | 3
e^x = 9 | ln
x = ln (9)

Should have isolated x first.

Anonymous

Anonymous Fri 02 Oct 2020 21:16:13 No.12187136 Report

Quoted By: >>12187246

>>12187007
Here are some which I still need to solve

e^x^2-2x = 2

e^x + 2 * e^-x =3

ln?x + 1,5 * lnx = ln(2x)

(lgx)^2 - lg x = 2

Anonymous

Anonymous Fri 02 Oct 2020 21:47:20 No.12187246 Report

Quoted By: >>12188995

>>12187136
Well, here's my process for equation 1:
$e^{x^{2}}-2x = 2$
There are two xs, one in an exponent, so if it's solvable it will probably involve lambert W.
That means I need to get it of the form $xe^x$
I want the terms with xs to multiply, so to do that' I'll move 2x to the RHS:
$e^{x^{2}} = 2+2x$
$e^{x^{2}} = 2(x+1)$
Now dividing:
$(x+1)^{-1}e^{x^{2}} = 2$
I don't think this can be isolated further because it is not of the lambert W form.
Even if a log is taken to get the x^2 term down, the other x term will be stuck in a log.

The second one is much simpler.
All x terms are in exponents, so you can just take logs:
$e^x + 2e^{-x} = 3$
We want all exponentials containing x to be multiplied so that we can take the logarithm, so we should get all exponential terms to be the same.
I will multiply both sides by $e^x$
$\left(e^{x}\right)^2 + 2 = 3e^x$
This leaves us with a quadratic equation if we replace e^x with u.
$\left(e^{x}\right)^2 - 3e^x + 2 = 0$
$u^2 - 3u + 2 = 0$
$(u-2)(u-1) = 0$
$u = 1, u = 2$
Substituting back in:
$e^x = 1, e^x = 2$
$x = ln(1), x = ln(2)$
$x = 0, x = ln(2)$
Now, it is necessary to check that both solutions are valid.
$e^0 + 2e^{-0} = 3$
$1 + 2 = 3$
$e^{ln(2)} + 2e^{-ln(2)} = 3$
$2 + 1 = 3$
They are both valid.

Anonymous

Anonymous Sat 03 Oct 2020 09:56:02 No.12188995 Report

Quoted By: >>12189303

>>12187246
Thank you!
So first you think about how to get x in a form which will be easy to solve later on in the context of the whole equation?

Anonymous

Anonymous Sat 03 Oct 2020 11:40:17 No.12189094 Report

Quoted By:

>>12187003
I'll finish off the other anon in case if you haven't done these already.
For the third:
ln(?x)+1.5*lnx = ln(2x)
Take out the square root to the front:
0.5lnx + 1.5lnx = ln(2x)
2lnx = ln(2x)
bring the two back up:
ln(x^2) = ln(2x)
as of now you can simply eyeball this, but formally you can take out the terms:
x^2 = 2x
x^2 - 2x = 0
x(x - 2) = 0
x = 0, 2
since ln0 is not possible, only 2 remains as a solution.

For 4: (assuming log means base 10 not exact form)
(logx)^2 - logx = 2
let logx = a
easy quadratic:
a^2 - a = 2
(a - 2)(a + 1) = 0
a = 2, -1
logx = 2, -1
x = 100, 1/10

Anonymous

Anonymous Sat 03 Oct 2020 12:58:27 No.12189207 Report

Quoted By:

>>12187089
That's because you should solve it as system of equation replacing logarythms with exponential form and not use e, if you're beginner

Anonymous

Anonymous Sat 03 Oct 2020 13:59:14 No.12189303 Report

Quoted By:

>>12188995
Yeah

Anonymous

Anonymous Sun 04 Oct 2020 03:27:09 No.12191330 Report

Quoted By:

>>12187003
I would rather hire a mathematician that solves problems differently than everyone else than anyone else.

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