Solve log[exp(x)+logx]=logx-exp(x) ??

1 Answer
Sep 16, 2017

# 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:

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So, we establish that there is a single solution, #0 lt x lt 1#, which we attempt to find numerically. We could use Newton-Rhapson but as this question is posed at precalculus level, let us instead use an iterative approach, by rearranging the equation into the form:

# x = g(x) # and use an iteration #x_(n+1) = g(x_n) #

There will many functions, #g(x)#, that we can choose, some may work and converge, others may not work or diverge.

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:

Steve M