Solve log[exp(x)+logx]=logx-exp(x) ??
1 Answer
# x = 0.28519 #
Explanation:
We have:
# log(exp(x)+logx) = logx-exp(x) #
Assuming that all logarithms are natural logarithms, then equivalently:
# ln(e^x + lnx) = lnx - e^x #
This equation cannot be solved analytically, so first we graph the functions to get a "feel" for the solutions:
So, we establish that there is a single solution,
# x = g(x) # and use an iteration#x_(n+1) = g(x_n) #
There will many functions,
Here we try the following iterative equation:
# ln(e^x + lnx) = lnx - e^x #
# :. e^x + lnx = e^(lnx - e^x) #
# :. lnx = e^(lnx - e^x) - e^x #
# :. x = e^(e^(lnx - e^x) - e^x) #
So we attempt
# x_n \ \ \ \ = 0.5 #
# x_(n+1) = exp(exp(lnx_n - exp(x_n)) - exp(x_n)) #
And using excel, working to 5dp we get: