3^x - 2^(x-1) = 3^(x-2) + 2^(x+1) 3^x - 3^(x-2) = 2^(x+1) + 2^(x-1) (3^x)(1-1/9) = (2^x)(2+1/2) (3/2)^x = (8/9) / (5/2) (3/2)^x = 45/16 Am I mistaken? Is there anything else to calculate the solution?

Question

3^x - 2^(x-1) = 3^(x-2) + 2^(x+1) <=> 3^x - 3^(x-2) = 2^(x+1) + 2^(x-1) <=> (3^x)(1-1/9) = (2^x)(2+1/2) <=> (3/2)^x = (8/9) / (5/2) <=> (3/2)^x = 45/16 Am I mistaken? Is there anything else to calculate the solution?

amistre64 · Answer

i assume your solving for x?

anonymous · Answer

Yes!

amistre64 · Answer

3^x - 2^(x-1) = 3^(x-2) + 2^(x+1) 

3^(x) - 2^(x) a = 3^(x) b + 2^(x) c 

3^(x) - 3^(x) b =  2^(x) c  + 2^(x) a

3^(x) (1-b) =  2^(x) (c+a)

logs help when working with exponents so ,

log[3^(x) (1-b)] =  log[2^(x) (c+a)]

log[3^(x)] + log[(1-b)] =  log[2^(x)] + log[(c+a)]

x log[3] + log[(1-b)] =  x log[2] + log[(c+a)]

x log[3] - x log[2] =  log[(c+a)] - log[(1-b)]

x ( log[3] - log[2]) =  log[(c+a)] - log[(1-b)]

thats my thoghts

anonymous · Answer

Yep, I see! I am supposed to do so without logarithms though as I will learn about them in the next chapter..

amistre64 · Answer

without logs?  then some deformed rt :)

amistre64 · Answer

maybe ...

3^(x) (1-b) =  2^(x) (c+a)

(3/2)^x  =  (c+a)/(1-b)

trial and error if the left side plays nice

amistre64 · Answer

a=1/2,  b = 1/9,  c = 2

3*9      3*3*3
--- =  -------
2*8     2*2*2*2


almost ....  as is i dont see a 'simple' solution since the number of 3s does not equal the number of 2s

anonymous · Answer

Thats what I ended up with too so I guess thats it..

amistre64 · Answer

thats the best i can do :)  good luck