Solve Exponential Equation 4^(5-9x)=1/(8^(x-2) )

[RidiculousName]

I'm trying to figure out how to solve 4^5-9x=1/(8^x-2) for x.

I've gotten as far as
2^2(5-9x) = 1/(2^3(x-2))
2(5-9x) = 1/(3(x-2))
10 - 18x = 1/(3x-6)
But, I'm not sure how to deal with the fraction.

EDIT: swapped 4^5-9x=1/(x^x-2) with 4^5-9x=1/(8^x-2)

 

[HOI]

I'm trying to figure out how to solve 4^5-9x=1/(x^x-2)

I've gotten as far as
2^2(5-9x) = 1/(2^3(x-2))

What happened to the x as a base in the exponential on the right?

2(5-9x) = 1/(3(x-2))
No! If a^x= 1/a^y it is NOT true that x= 1/y. a^x= 1/a^y= a^{-y} so that x= -y

[tex]10 - 18x = 1/(3x-6)
But, I'm not sure how to deal with the fraction.

IF that were really the equation you needed to solve, you would multiply both sides by 3x- 6 to get a quadratic equation.

HOWEVER, you have several mistakes!

If the problem really is [tex]4^{5- 9x}= \frac{1}{x^{x- 2}}[/tex] then the equation can only be solved using the "Lambert W function" (defined as the inverse function to [tex]f(x)= x^x[/tex]).

IF "x" in the base was a typo and the problem was [tex]4^{5- 9x}= \frac{1}{8^{x- 2}}[/tex], THEN we can write it as [tex](2^2)^{5- 9x}= (2^3)^{-(x- 2)}[/tex].

[tex]2^{2(5- 9x)}= 2^{-3(x- 2)}[/tex].

[tex]2^{10- 18x}= 2^{-3x+ 6}[/tex].

Now, taking the logarithm of both sides, or just using the fact that [tex]2^x[/tex] is a "one to one" function, we have 10- 18x= -3x+ 6.

 

[RidiculousName]

My mistake!
4^5-9x=1/(x^x-2) should be 4^5-9x=1/(8^x-2)
Sorry!

 

[Wilmer]

SO (1st step): 4^(5-9x) = 8^(2-x)

 

[RidiculousName]

SO (1st step): 4^(5-9x) = 8^(2-x)

Almost. It's 4^(5-9x) = 1/8^(x-2) not 4^(5-9x) = 8^(2-x)

 

[Wilmer]

Almost. It's 4^(5-9x) = 1/8^(x-2) not 4^(5-9x) = 8^(2-x)

I'm trying to tell you that IF:
4^(5-9x) = 1/8^(x-2)
THEN:
4^(5-9x) = 8^(2-x)

If you do not understand that 1/8^(x-2) = 8^(2-x),
then you are not ready for this problem.

Still confused? Look-see here:
Wolfram|Alpha: Computational Intelligence)

RULE: 1/n^(-p) = n^p

 

[RidiculousName]

If you do not understand that 1/8^(x-2) = 8^(2-x),
then you are not ready for this problem.

If you feel forgetting one piece of a puzzle means you aren't ready to learn it at all then, respectfully, that's your own issue. Thank you for the help anyway.

 

[Greg]

$$4^{5-9x}=\left(\frac18\right)^{x-2}=8^{2-x}=2^{3(2-x)}$$

$$2^{2(5-9x)}=2^{3(2-x)}$$

$$2(5-9x)=3(2-x)$$

$$10-18x=6-3x$$

$$4=15x$$

$$x=\frac{4}{15}$$