Can't figure out how to solve by grouping

[appearance]

Hey, I can't figure out how to group this equation or solve it by any other method. I would appriciate a step-by-step guide because I'm lost.

5^3x + 9 * 5^x + 27(5^-3x + 5^-x) = 64

The anwsers should be 0 and log5(3).

Edit: Sure it was +9 * 5^x, my bad!

 

[tkhunny]

What have you tried? Can you rewrite all the \(\displaystyle 5^{something}\) terms into a similar form?

 

[appearance]

I've tried substitution and multiplying the brackets and regrouping the terms. I can't solve it by any means. I would appriciate some help!

 

[Deleted member 4993]

I've tried substitution and multiplying the brackets and regrouping the terms. I can't solve it by any means. I would appriciate some help!

What type of substitution did you try?

Please share your work step-by-step.

 

[appearance]

By expanding the brackets i got:

5^3x + 9 * 5^x + 27 * 5^-3x + 27 * 5^-x = 64 | ( * 5^-3 )

Now I've got:

5^6x + 9 * 5^4x + 27 + 27 * 5^2x = 64 * 5^3x

Now making it prettier:

5^6x + (9 * 5^4x) - (64 * 5^3x) + (27 * 5^2x) + 27= 0

At this point I don't know how to regroup them

I tried substition, but it was really hopeless. I substituted 5^x, but the different powers didn't allow to do anything further.

5^x = a

a⁶ + 9a⁴ - 64a³ + 27a² + 27= 0

 

[Deleted member 4993]

By expanding the brackets i got:

5^3x + 9 * 5^x + 27 * 5^-3x + 27 * 5^-x = 64 | ( * 5^-3 )

Now I've got:

5^6x + 9 * 5^4x + 27 + 27 * 5^2x = 64 * 5^3x

Now making it prettier:

5^6x + (9 * 5^4x) - (64 * 5^3x) + (27 * 5^2x) + 27= 0

At this point I don't know how to regroup them

I tried substition, but it was really hopeless. I substituted 5^x, but the different powers didn't allow to do anything further.

5^x = a

a⁶ + 9a⁴ - 64a³ + 27a² + 27= 0

Applying rational root theorem we find that "1" is one of the rational roots of the derived polynomial.

a⁶ + 9a⁴ - 64a³ + 27a² + 27 = (a - 1)(a⁵ + a⁴ + 10a³ - 54a² - 27a - 27)

 

[Deleted member 4993]

Hey, I can't figure out how to group this equation or solve it by any other method. I would appriciate a step-by-step guide because I'm lost.

5^3x - 9 * 5^x + 27(5^-3x + 5^-x) = 64

The anwsers should be 0 and log5(3).

At x = 0,

5^3x - 9 * 5^x + 27(5^-3x + 5^-x) = 5^3*0 - 9 * 5^*0 + 27(5^-3*0 + 5^-0) = 1 - 9 + 27(1 + 1) \(\displaystyle \ne\)64

are you sure it is not:

5^3x + 9 * 5^x + 27(5^-3x + 5^-x) = 64

 

[appearance]

At x = 0,

5^3x - 9 * 5^x + 27(5^-3x + 5^-x) = 5^3*0 - 9 * 5^*0 + 27(5^-3*0 + 5^-0) = 1 - 9 + 27(1 + 1) \(\displaystyle \ne\)64

are you sure it is not:

5^3x + 9 * 5^x + 27(5^-3x + 5^-x) = 64

Indeed it is...

Any idea how to get the second root?

Thank you so much for the first one!

 

[Deleted member 4993]

Indeed it is...

Any idea how to get the second root? ..... The same way I showed 5^x = 1 is a solution!

Thank you so much for the first one!

.