6*(9^(1/x))-13*(6^(1/x))+6*(4^(1/x))=0
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.
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?
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.
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 .
To find the value of x that satisfies the equation 6*(9^(1/x))-13*(6^(1/x))+6*(4^(1/x))=0, we can use a method called substitution:
Step 1: Let's substitute a variable to make calculations easier. We will use u = 9^(1/x), v = 6^(1/x), and w = 4^(1/x). Now, we can rewrite the equation as 6u - 13v + 6w = 0.
Step 2: Raise each side of the equation to the power of x in order to get rid of the fractional exponents. We will have (6u - 13v + 6w)^x = 0.
Step 3: Since any number, when raised to power 0, equals 1, we have (6u - 13v + 6w)^x = 0 for any value of x, except when 6u - 13v + 6w = 0.
Step 4: Now, let's substitute back the values of u, v, and w that we initially set. We should solve the equation 6*(9^(1/x))-13*(6^(1/x))+6*(4^(1/x)) = 0:
6*(9^(1/x)) - 13*(6^(1/x)) + 6*(4^(1/x)) = 0
6*(9^(1/x)) - 13*(6^(1/x)) = -6*(4^(1/x))
(6/4)*3*(9^(1/x)) - (13/4)*3*(6^(1/x)) = -6*(4^(1/x))
(3/2)^2*(9^(1/x)) - (13/2)*(3/2)*(6^(1/x)) = -6*(4^(1/x))
(3/2)^2*(9^(1/x)) - (13/2)*(3/2)*(6^(1/x)) + (13/2)^2*(6^(1/x))^2 = -6*(4^(1/x)) + (13/2)^2*(6^(1/x))^2
(3/2)^2*(9^(1/x)) - (13/2)*(3/2)*(6^(1/x)) + (13/2)^2*(6^(1/x))^2 = -6*(4^(1/x)) + (13/2)^2*(6^(1/x))^2
Now, let's substitute y = (6^(1/x)):
(3/2)^2*(9^(1/x)) - (13/2)*(3/2)*y + (13/2)^2*y^2 = -6*(4^(1/x)) + (13/2)^2*y^2
(3/2)^2*(9^(1/x)) - (13/2)*(3/2)*y + (13/2)^2*y^2 + 6*(4^(1/x)) = (13/2)^2*y^2
(3/2)^2*(9^(1/x)) - (13/2)*(3/2)*y + (13/2)^2*y^2 - (13/2)^2*y^2 = -6*(4^(1/x))
(3/2)^2*(9^(1/x)) - (13/2)*(3/2)*y - (13/2)^2*y^2 + (13/2)^2*y^2 = -6*(4^(1/x))
Now, we simplify:
(3/2)^2*(9^(1/x)) - (13/2)*(3/2)*y = -6*(4^(1/x))
At this point, we can plug the values of 9^(1/x) and 6^(1/x) back in and solve for y. However, without knowing the specific values of 9^(1/x) and 6^(1/x), we cannot further simplify the equation algebraically.
To find the value of y, you will need to use numerical methods such as approximation or numerical solvers like Newton-Raphson method or iteration techniques.