Question

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

Answer 1

`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, `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: