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/sci/ - Science & Math

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10470255 No.10470255 [Reply] [Original]

e^x + e^-x = 5

I've had this question on my math test today, and couldn't solve at all. Somebody can show my stupid ass how to solve it?

>> No.10470261

e^x + 1/e^x - 5 = 0

Figure it out from there.

>> No.10470271

>>10470261
Ive tried to write it in form of y^2 - 5y + 1 = 0, where y = e^x
But I couldn't get any further

>> No.10470282

>>10470271
Use quadratic equation to solve for y

>> No.10470287

>>10470271
quadratic formula

>> No.10470288

e^x=(5+/-sqrt(21))/2
x = ln((5+/-sqrt(21))/2)

>> No.10470290

>>10470282
Ok, got it. Thanks! Is this the only way tho?

>> No.10470292

You could also use the cosh^-1 formula

>> No.10470296

>>10470292
haven't leaved it yet

>> No.10470301

>>10470290
You could complete the square instead of using the quadratic equation if you really wanted to.

>> No.10470302

>>10470296
Learned*

>> No.10470311
File: 7 KB, 272x154, 1543396474179.png [View same] [iqdb] [saucenao] [google]
10470311

>>10470296
cosh(x) is the same as pic related

>> No.10470312

>>10470301
I tried to do exactly that, forgetting about quadratic equation, but couldn't find right factors

>> No.10470315

>>10470311
So, if e^x+e^-x=5, cosh(x)=5/2 and x = cosh^-1(5/2)

>> No.10470326

>>10470315
I see what u did here, but what is the next step to find exact value for x?
I don't quite know how to use 'cosh' function

>> No.10470341

>>10470312
That's why I said if you really wanted to, using the quadratic equation is far less painful.

>> No.10470372

I replaced e^x with a generic number "a" (could be y, or any letter you want), and I transformed it into a quadratic equation.

a^2 - 5a + 1 = 0

You'll find
a1=(5+√21)/2
and
a2=(5-√21)/2
Therefore x=ln((5+√21)/2)
or x=ln((5+√21)/2)

That's how I'd do it.

>> No.10470379

>>10470372
**or x=ln((5-√21)/2)

>> No.10470399

>>10470372
Now, finishing with a calculator, I found x=~1,566 or x=~-1,566.

Did you have a calculator on you? If not, than you were probably supposed to give the answer in the "x=ln(blabla)" form

>> No.10470435

>>10470255
Hint: Replace e^x with y.

>> No.10471642

>>10470255
Absolute state of /sci/