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

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I'm trying to figure out how to solve 45-9x=1/(8x-2) for x.

I've gotten as far as
22(5-9x) = 1/(23(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 45-9x=1/(xx-2) with 45-9x=1/(8x-2)
 
Last edited:
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RidiculousName said:
I'm trying to figure out how to solve 45-9x=1/(xx-2)

I've gotten as far as
22(5-9x) = 1/(23(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

10 - 18x = 1/(3x-6)<br /> But, I&#039;m not sure how to deal with the fraction.
<br /> IF that were really the equation you needed to solve, you would multiply both sides by 3x- 6 to get a quadratic equation.<br /> <br /> HOWEVER, you have several mistakes!<br /> <br /> If the problem really is 4^{5- 9x}= \frac{1}{x^{x- 2}} then the equation can only be solved using the &quot;Lambert W function&quot; (defined as the inverse function to f(x)= x^x).<br /> <br /> IF &quot;x&quot; in the base was a typo and the problem was 4^{5- 9x}= \frac{1}{8^{x- 2}}, THEN we can write it as (2^2)^{5- 9x}= (2^3)^{-(x- 2)}.<br /> <br /> 2^{2(5- 9x)}= 2^{-3(x- 2)}. <br /> <br /> 2^{10- 18x}= 2^{-3x+ 6}.<br /> <br /> Now, taking the logarithm of both sides, or just using the fact that 2^x is a &quot;one to one&quot; function, we have 10- 18x= -3x+ 6.
 
My mistake!
45-9x=1/(xx-2) should be 45-9x=1/(8x-2)
Sorry!
 
SO (1st step): 4^(5-9x) = 8^(2-x)
 
Wilmer said:
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)
 
RidiculousName said:
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
 
Last edited:
Wilmer said:
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.
 
$$4^{5-9x}=\left(\frac18\right)^{x-2}=8^{2-x}=2^{3(2-x)}$$

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

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

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

$$4=15x$$

$$x=\frac{4}{15}$$
 

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