Let's Solve A Fun Exponential Equation

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SyberMath

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Let u=sqrt(x)
Given equation becomes
(3^u)+(6^u)=(12^u)
(3^u)+(2^u)(3^u)=(2^u)²(3^u)
As 3^u≠0 we can divide by 3^u
(2^u)²-(2^u)-1=0
2^u is golden ratio, say ß
2^u=ß>0
Take log based 2: u=²log(ß)
x=[²log(ß)]²
where ß=½[1+sqrt(5)]

nasrullahhusnan

enjoyed solving this.Like out of the book stuff

sunildhuri

If you divide by 3^sqr(x) then you get powers of 2 instead of powers of 1/2. You could divide by the power of 6 also, its just a little bit more complicated.

DonRedmond-jkhj

You should give an approximate value, which is x = 0.48197, so people can better compare their results to yours.

goldfing

let y = sqrt(x) >=0
write 6 and 12 in term of their prime decomposition.
divide both sides by 2^y*3^y
you get an equation only involving X=2^y. X>=1>0 given y>=0
it reduces as X^2-X-1=0
solution = p = golden ratio = (1+sqrt(5))/2, the other solution is negative=> not possible given X>0.
x=2^y=p
y=ln(p)/ln(2)
x=[ (ln(p)/ln(2) ]^2

Fred-yqfs