What is the answer for a^x + a^x? when a^2x + a^-2x = 6, a > 0, and 0 < a^x < 1.

[Jihyun Kim]

I am studying Exponential equations now. I don't know if this problem is related to those subjects.

What is the answer for a^x + a^x? when a^2x + a^-2x = 6, a > 0, and 0 < a^x < 1.​

Please, give me advice.

Thank you.

 

[BigBeachBanana]

I used latex to rewrite what I believe the question is, can you confirm? Also, what have you tried so far?
Find [imath]a^x+a^x[/imath],
when [imath]a^{2x}+a^{-2x}=6[/imath], [imath]a>0[/imath], and [imath]0< a^x<1[/imath]

 

[Deleted member 4993]

I am studying Exponential equations now. I don't know if this problem is related to those subjects.
Please, give me advice.

Thank you.

You wrote:

What is the answer for a^x + a^x? when a^2x + a^-2x = 6, a > 0, and 0 < a^x < 1.

Could it be:

What is the answer for a^x + a^(-x)? when a^2x + a^-2x = 6, a > 0, and 0 < a^x < 1.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem.

 

[JeffM]

Exponential equations that involve addition or subtraction may be very hard or impossible to solve exactly. Here a substitution will work.

[math]\text {Let } b = a^{2x}.[/math]
[math]b + \dfrac{1}{b} = 6 = \dfrac{6b}{b} \implies b = \dfrac{6b - 1}{b}.[/math]
Solve for b and proceed.

 

[Steven G]

Exponential equations that involve addition or subtraction may be very hard or impossible to solve exactly. Here a substitution will work.

[math]\text {Let } b = a^{2x}.[/math]
[math]b + \dfrac{1}{b} = 6 = \dfrac{6b}{b} \implies b = \dfrac{6b - 1}{b}.[/math]
Solve for b and proceed.

Jeff,
I respect you an enormous amount but this time I feel that maybe you were tired or drunk (probably both) when you wrote your reply. Although it is 100% correct, it is not the cleanest, clearest way to solve for b.
To solve for b+ 1/b = 6, just multiply both sides by b to get b^2 + 1 = 6b....

 

[Steven G]

Can you at least simplify a^x + a^x?

 

[JeffM]

Jeff,
I respect you an enormous amount but this time I feel that maybe you were tired or drunk (probably both) when you wrote your reply. Although it is 100% correct, it is not the cleanest, clearest way to solve for b.
To solve for b+ 1/b = 6, just multiply both sides by b to get b^2 + 1 = 6b....

Must have been tired. Seldom drink before noon.

You are right. What I saw as the cleanest way is not the cleanest. Was focused on the idea of substitution.

 

[lookagain]

What is the answer for a^x + a^x? when a^2x + a^-2x = 6, a > 0, and 0 < a^x < 1.​

Please, give me advice.

Advice: When you write that equation with those relatively more complicated exponents as you did here in horizontal style, you must enclose those exponents because of the Order of Operations. a^(2x) + a^(-2x) = 6

Also, Subhotosh Khan asked the good question if it should actually be for a^x + a^(-x).

 

[Dr.Peterson]

I am studying Exponential equations now. I don't know if this problem is related to those subjects.

What is the answer for a^x + a^x? when a^2x + a^-2x = 6, a > 0, and 0 < a^x < 1.​

Please, give me advice.

Thank you.

If there is a typo there, and the problem is really

Find [imath]a^x+a^{-x}[/imath], given that [imath]a^{2x}+a^{-2x}=6[/imath], where [imath]a>0[/imath], and [imath]0< a^x<1[/imath]​

then there is a very nice way to solve it. See what you get when you square [imath]a^x+a^{-x}[/imath].

But the last condition in the problem makes me suspect this is not what the problem is.