Are all the logs to base 10?
Do you mean 9^(log3*logx) or 9^log3 * logx ?
And does log^2x mean (logx)^2 ?
One way of reading your equation is
9^log3 logx = logx - 2(logx)^2 + 4
which is just a quadratic in logx:
2(logx)^2 + (9^log3 - 1) logx - 4 = 0
so, if you let u=logx, then that is just
2u^2 + 1.853u - 4 = 0
which is easy to solve.
Somehow I think you have mangled the equation.
9^log3logx = log x -2log^2x + 4
2 answers
Assuming you meant:
(9^log3)logx = log x - 2log^2 x + 4
let logx = k , then you have
(9^log3)k = k - 2k^2 + 4
2k^2 + k(9^log3 - 1) - 4 = 0
looks like some calculator work ahead
use the quadratic formula where
a = 2 , b = 9^log3, and c = -4
to find k
remember k = logx , then find x
(9^log3)logx = log x - 2log^2 x + 4
let logx = k , then you have
(9^log3)k = k - 2k^2 + 4
2k^2 + k(9^log3 - 1) - 4 = 0
looks like some calculator work ahead
use the quadratic formula where
a = 2 , b = 9^log3, and c = -4
to find k
remember k = logx , then find x
0 answers