If you mean
(9^x)^(-2) + (3^3)^x = 81
then you have
3^(-4x) + 3^(3x) = 81
if u = 3^x, then you have
1/u^4 + u^3 = 81
I'm sorry, but I don't see any algebraic solution. you sure there's no typo? Wolframalpha gets this:
http://www.wolframalpha.com/input/?i=3^%28-4x%29+%2B+3^%283x%29+%3D+81
9^x(-2)+3^3(x)=81 i need help
5 answers
steve can logarithm work 4 dat
I did similar what Steve did, and agree with him
Steve asked you to improve your typing of the equation, but this does not change things.
I will assume you meant:
9^(-2x) + 3^(3x) = 81
which becomes
3^(-4x) + 3^(3x) = 81
(3^-x)^4 + (3^-x)^-3 = 81
let 3^-x = t
so we have:
t^4+ t^-3 = 81
times t^3 for everybody
t^7 + 1 = 81t^3
tough to solve, unless we use something like Wolfram
http://www.wolframalpha.com/input/?i=t%5E7+%2B+1+%3D+81t%5E3
notice t = appr 3 or t = appr -3 are solutions to our last equation,
so 3^-x = 3 = 3^1
x = -1
similarly x = appr +1
somehow I don't think that is what you meant.
Steve asked you to improve your typing of the equation, but this does not change things.
I will assume you meant:
9^(-2x) + 3^(3x) = 81
which becomes
3^(-4x) + 3^(3x) = 81
(3^-x)^4 + (3^-x)^-3 = 81
let 3^-x = t
so we have:
t^4+ t^-3 = 81
times t^3 for everybody
t^7 + 1 = 81t^3
tough to solve, unless we use something like Wolfram
http://www.wolframalpha.com/input/?i=t%5E7+%2B+1+%3D+81t%5E3
notice t = appr 3 or t = appr -3 are solutions to our last equation,
so 3^-x = 3 = 3^1
x = -1
similarly x = appr +1
somehow I don't think that is what you meant.
yes reiny daz the question i jus decided to include bracket in it,reiny u got me confuse how come 3^x become 3^-x,but what if i use logarithm can it work?
You must be referring to the 2nd term of
(3^-x)^4 + (3^-x)^-3 = 81
notice in the previous line it was 3^(3x)
which is (3^x)^3 or (3^(-x))^-3
Steve substituted 3^x, I substituted 3^-x,
our equations reflect that
No, logs will not work here, since you have to take the log of a sum, and we have no formulas to take care of log (A + B)
(3^-x)^4 + (3^-x)^-3 = 81
notice in the previous line it was 3^(3x)
which is (3^x)^3 or (3^(-x))^-3
Steve substituted 3^x, I substituted 3^-x,
our equations reflect that
No, logs will not work here, since you have to take the log of a sum, and we have no formulas to take care of log (A + B)