1 = logx(a) * logx(b)

[Relmiw]

Would someone please help me in solving for x in the following equation?

1 = log_x(a) * log_x(b)

Thank you very much for any help

 

[pka]

Would someone please help me in solving for x in the following equation?
1 = log_x(a) * log_x(b)

Your post count indicates that you have at least four posts. So you should know how it works here.
Therefore, where is your own work?

Hint: \(\displaystyle {\Large c^{\log_c(p)}=p}\)

 

[Relmiw]

Your post count indicates that you have at least four posts. So you should know how it works here.
Therefore, where is your own work?

Hint: \(\displaystyle {\Large c^{\log_c(p)}=p}\)

Sorry, it's been a few years since my other thread. The posted equation is the result of my work and is the phase at which I'm stuck. I'll back up a couple steps to show how I've arrived here.

x^(1/u) = a
x^u = b

logx(a) = 1/u
logx(b) = u

logx(a)*logx(b) = 1

I'm aware there may be a different way to go about solving the original system of equations, but i think there is also a way to solve for x in terms of a and b in the final equation.

Thank you for any help, unfortunately I am having trouble making progress from your hint.

 

[stapel]

I'll back up a couple steps to show how I've arrived here.

x^(1/u) = a
x^u = b

Is x even relevant?

. . . . .\(\displaystyle x^{\frac{1}{u}}\, =\, \sqrt{x\,}\, =\, a\)

. . . . .\(\displaystyle x\, =\, a^u\)

. . . . .\(\displaystyle x^u = \left(a^u\right)^u\, =\, a^{u^2}\, =\, b\)

Now solve for "a" in terms of "b". (I'm assuming you know the value of "u".) You can back-solve for x, if needed.

 

[Relmiw]

I thought someone would be able to look at this and come up with a slick way to isolate x in terms of a and b, but maybe it's not as simple as I at first thought.

I do not know the value of u (though I see how I can find the value of two of these given the other two). I'm more interested in a general result resembling x = something something "a" something something "b"

Thank you both for your contributions

 

[stapel]

I do not know the value of u....

That would have been useful information. Lacking the value of "u", you're stuck with the hint provided in the initial reply.

I thought someone would be able to look at this and come up with a slick way to isolate x in terms of a and b, but maybe it's not as simple as I at first thought.

What did you get when you used that first hint?

Please be complete. Thank you!

 

[Deleted member 4993]

Sorry, it's been a few years since my other thread. The posted equation is the result of my work and is the phase at which I'm stuck. I'll back up a couple steps to show how I've arrived here.

x^(1/u) = a
x^u = b

logx(a) = 1/u
logx(b) = u

logx(a)*logx(b) = 1

I'm aware there may be a different way to go about solving the original system of equations, but i think there is also a way to solve for x in terms of a and b in the final equation.

Thank you for any help, unfortunately I am having trouble making progress from your hint.

(1/u)*ln(x) = ln(a)

u * ln(x) = ln(b)

[ln(x)]² = ln(a)*ln(b)

ln(x) = [ln(a)*ln(b)]^1/2

finish it......

 

[Relmiw]

That would have been useful information. Lacking the value of "u", you're stuck with the hint provided in the initial reply.


What did you get when you used that first hint?

Please be complete. Thank you!

Whether or not I have a value for "u" does not make a difference for the question I posed to you. My first post is complete.

(1/u)*ln(x) = ln(a)

u * ln(x) = ln(b)

[ln(x)]² = ln(a)*ln(b)

ln(x) = [ln(a)*ln(b)]^1/2

finish it......

Thank you very much Subhotosh Khan, that is exactly what I was looking for.