Solving for unknown - natural log involved

**danidh325** · November 10th 2008, 01:24 PM

Hello.

I am a bit rusty on my natural log rules. How would I go about solving for x:

.25ln(11000 + x) + .75ln(10000 + x) = .25ln(11000) + .75ln(10000)

Any help would be greatly appreciated.

**Plato** · November 10th 2008, 02:18 PM

If you had $a\ln (b + x) + c\ln (d + x) = a\ln (b) + c\ln (d)$ then it follows that:
$\begin{gathered}<br /> \ln \left[ {(b + x)^a (d + x)^c } \right] = \ln \left[ {(b)^a (d)^c } \right] \hfill \\<br /> (b + x)^a (d + x)^c = (b)^a (d)^c \hfill \\ <br /> \end{gathered}$ .
That at least get it down to x’s.

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