How do, /sci/?
Isolate a single variable and solve, then go back and solve for the others. Literally how hard is this?
>>7918190
Only z can be found that way.
>>7918192
Find z, you have it in integers. use it as a solution to the other equation, assuming they belong to the same system. (YOU MIGHT WANT TO USE A LOGARITHM)
>>7918192
Solve for Z.
Divide out the Y in the left equation.
Use quadratic equation
That is now equal to the left equation
Isolate your variable of choice
Solve
Of course I'm a retard so this is probably wrong :^)
>>7918200
Yeah, I found z.
I put it into the first equation.
I still have
[math]
(z+4)5^{|y|-1} \\
\log{(z+4)}+\log{5^{|y-1|}} \\
\log{z+4}+(|y|-1)\log{5} \\ = (x^2+4xy+3)\log{6}
can't simplify from there and isolate one
[\math]
>>7918222
oops
[math]
(z+4)5^{|y|-1} \\
\log{(z+4)}+\log{5^{|y-1|}} \\
\log{(z+4)}+(|y|-1)\log{5} = (x^2+4xy+3)\log{6}
[/math]
can't simplify from there and isolate one
>>7918228
Can you factorise the polynomial?
>>7918233
What? No you can't
>>7918234
what is the quadratic formula?
>>7918234
[math]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/math]
:*
>>7918265
Yes. Now go to fucking stack exchange when you want homework help.
>>7918267
can you just help
>>7918267
What are your integer solutions for z?
>>7918265
[eqn]x=\frac{-4y\pm\sqrt{(16y^2-12)}}{2}[/eqn]
useless
>>7918273
[math]
z=4n+3 where n is an integer
>>7918281
Ok, now you've simplified the top equation into one of two variables.
6^(Some integer)=(4n+7)*5^(some integer)
You now need to find x and y such that the answers are multiples of 5 and 6. This is relatively easy through brute force, but there's a better method
>>7918286
Yeah, I know I can do it this way.
I'm looking for the "better" method.
>>7918288
Find the greatest common divisor of 6^n and (4n+7)*5^m.
Have you done any discrete mathematics?
>>7918302
Either my python code is shit or there's no solution to this.
>>7918339
You tried graphing them?
5 divides the right side of the first equation. How can it divide the left? There are only a couple possibilities here.
>>7918355
5 is prime
>>7918355
This is a very big hint. There might not be a series of solutions for x and y, just one or two.
>>7918355
The second equation tells us z=(4 n+1)/3, where n is an integer. The
left side of the first equation is positive. So the right size is too.
Thus z > -4. If |y|>1, then the right size is divisible by 5, which the
left can't be, since the left's prime divisors are 2 and 3. So y=0 or
y=1 or y=-1. If y=0, then 5*6^(x^2+3)-4 = z = (4n+1)/3, or
3*5*6^(x^2+3)-13=4n. The right is divisible by 4, the left not. So y=0
is impossible.
Two more cases, y=1, y=-1...
>>7918386
Why -4? (5*6^(x^2+3)-4 = z = (4n+1)/3))
hijacking this anon's thread to ask for help for an easier thing. - sorry op
I established u as sent and dv = e^-(s+1)t dt, but Im not getting how Im supposed to get it (as the highlighted) any help would be appreciated
>>7918415
why
sin( 3 z pi /2 ) = 1, so 3 z pi /2 = pi/2 + 2 pi n, (n an integer). So 3 z pi = pi + 4 pi n, so 3 z = 1 + 4 n, so z = (4 n + 1) / 3. But z must
be an integer. The diophantine equation solution is z= 4m +3 , n=3m+2, where m is any integer. We just need to solve 6^(x^2+4xy+3) = (4 m +7)
5^(|y|-1). If y=0, then 5*6^(x^2+4xy+3) = (4 m +7) is impossible since 4 divides the left but not the right. If y=1 and x is not in {-3,-2,-1}, then x^2+4xy+3>2 so again 4 divides the left but not the right. Checking y=1 and x in {-3,-2,-1} yields no solutions. If y=-1,
and x is not in {1,2,3} then x^2+4xy+3>2, impossible. Checking y=-1 and x in {1,2,3} gives no solutions.
None.
Ugly, but none.
>>7918415
Maybe the integrate by parts twice trick?
>>7918463
yep, that was it. If anyone curious:
>>7918460
Why is the left divisible by 4?
>>7918960
6^2 is divisible by 4 => 6^n (n>=2) is divisible by 4
