Question

6*(9^(1/x))-13*(6^(1/x))+6*(4^(1/x))=0

Answer 1

You have powers of 2 and 3. Expand all the groups, collect the powers, and solve for x.

If you get stuck, show what you got done.

Answer 2

I understand how to do it but am having trouble with the mathematical aspect. I have redone the problem quite a few times but keep getting stuck. Could you help?

Answer 3

It's annoying how you don't want to show your work. Just let us take care of it all.

Just for readability, try letting u = 1/x. Then you have

6*(9^u)-13*(6^u)+6*(4^u))=0
6*3^(2u) - 13*3^u*2^u + 6*2^(2u)

Now if you let v=3^u and z=2^u you have

6v^2 - 13vz + 6z^2 = 0
(3v-2z)(2v-3z) = 0

3v = 2z
3^(u+1) = 2^(u+1)
true only if u+1=0, or u=-1 so, 1/x = -1, x = -1

2v = 3z
2*3^u = 3*2^u
3^(u-1) = 2^(u-1)
true only if u-1=0 or u=1, so x=1

So, the two solutions are x = -1, 1

Sometimes you just have to make things easier to read to wade through the noise.

Answer 4

let assume that 1/x ( it is easy for me to write it down)
6*(9^(1/x))-13*(6^(1/x))+6*(4^(1/x))=0
turn out : 6*9^a-13*6^a+6*4a^a=0
divide 9^a to the function
6-13*(2/3)^a +6*(4/9)^a=0 ([2/3]^2=4/9)
=> 6-13*(2/3)^a + 6(2/3)^2a=0
now you can see out function looks like : ax^2+bx+c=0 so just put it into the calculator and we ge two solotion
first :2/3^a = 2/3 <=>a=1 <=>1/x=1 => x=1
second: 2/3^a =3/2<=>a=-1 <=> x=-1

in this problem, you just need to divide for 9^(1/x) or 4^(1/x) and it turn out the function ax^2 +bx+c .