A Homemade Exponential System

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[raystinger]

So we have:
a^b = 2^(1/4)
Right off the bat it would be so great IF WE HAD a=2 and b=1/4. So let's try those values on the other equation:
a^(b³+b) = 2^(17/64) -> false
So (a, b)=(2, 1/4) isn't a solution, but maybe we can work around that:
Let's assume a=2^k and b=1/4k. That way we still have a^b = 2^(1/4), so we just have to work on the other equation:
a^(b³+b) = 2^((16k²+1)/(64k²)) = 2^(5/16)
=>(16k²+1)/(64k²) = 5/16
=>k = ±1/2
So our solution is:
(a, b) = (2^(±1/2), ±1/2)

[GirishManjunathMusic]

Given:
a↑(b³ + b) = +(32↑1/16) — ①
a↑b = 2↑¼ — ②
To find:
a and b:
Using x↑(y + z) = (x↑y)·(x↑z):
(a↑b³)·(a↑b) = 32↑1/16

Substituting in ②:
(2↑¼)·(a↑b³) = (2↑5)↑1/16

Using (x↑y)↑z = x↑yz:
(2↑¼)·(a↑b³) = 2↑5/16

Dividing both sides by 2↑¼:
a↑b³ = (2↑5/16)/(2↑¼)

Using (x↑y)/(x↑z) = x↑(y – z) and ¼ = 4/16:
a↑b³ = 2↑1/16

Using x↑y³ = (x↑y)↑y² and substituting in ②:
(2↑¼)↑b² = 2↑1/16

Using (x↑y)↑z = x↑yz:
2↑(¼b²) = 2↑1/16

Comparing exponents:
¼b² = 1/16

Multipying both sides by 4:
b² = 4/16
b² = ¼
b = ±½.

Substituting into :
When b = ½
a↑½ = 2↑¼

Squaring both sides:
a = 2↑½.

When b = –½
a↑(–½) = 2↑¼

Squaring both sides:
a↑(–1) = 2↑½

Inverting both sides:
a = 1/(2↑½)

Multiplying RHS by (2↑½)/(2↑½):
a = (2↑½)/2.

So:
(a, b) = (2↑½, ½) and ((2↑½)/2, –½)

[jmart]

I solved it taking logs. You obtain a very easy system of eqs. Interesting problem. Thanks.

[pwmiles]

It's really the same method but I took logs base 2
(b^3+b) log2(a) = 5/16
b log2(a) = 1/4

Divide on each side
b^2 + 1 = 5/4
b = +/- 1/2
a = sqrt(2) or 1/sqrt(2)

[hazalouldi]

Please, what is the value integers for :(x+y)^3=(x-y-6)^2 thanks

[MichaelJamesActually]

So you're telling me roots of a number can be negative...