Solve the function
2(9^x) - 5(3^×) = 12
So what I did So far is changed it into log functions:
2(9^x) = 12
9^x = 6
Log9(6) = x
And then
5(3^x) = 12
3^x = 12/5
Log3(12/5) = x
So then I don't know how to subtract them because they have different bases :/
4 answers
who are you and why are you using my name?
... my name's also rose? I'm just trying to get math help man
kay
2*(3^2)^x - 5*3^x = 12
2*3^x *3^x - 5*3^x = 12
let z = 3^x
2 z^2 -5 z -12 = 0
z = 4 or -3/2
so
3^x = 4
or
3^x = -1.5
now x log 3 = log 4 etc
2*3^x *3^x - 5*3^x = 12
let z = 3^x
2 z^2 -5 z -12 = 0
z = 4 or -3/2
so
3^x = 4
or
3^x = -1.5
now x log 3 = log 4 etc
1 answer