# How to solve q = ln &#x2061;<!-- ⁡ --> <mrow class="MJX-TeXAto

How to solve $q=\frac{\mathrm{ln}n}{\mathrm{ln}b+\mathrm{ln}q+\mathrm{ln}\mathrm{ln}n}$
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Janiyah Patton
Let $q={e}^{x}$ so that $\mathrm{ln}q=x.$. Also, put $A=\mathrm{ln}n,$, $B=\mathrm{ln}b,$, and $C=\mathrm{ln}\mathrm{ln}n.$. Then your equation takes the form
${e}^{x}\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}\frac{A}{B+x+C}$
Multiplying both sides by $B+x+C$ gives
${e}^{x}\left(B+x+C\right)\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}A$
Now make the variable change $u=B+x+C,$, so that $x=u-B-C.$.Then we get
${e}^{u-B-C}\cdot u\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}A$
Multiplying both sides by ${e}^{B+C}$ gives
${e}^{u}\cdot u\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}A{e}^{B+C}$
Now we can express the solution in terms of the Lambert function:
$u=W\left(A{e}^{B+C}\right)$
Hence, recalling that $u=B+x+C$ and $q={e}^{x}$ we get
$B+x+C\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}W\left(A{e}^{B+C}\right)$
$x\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}W\left(A{e}^{B+C}\right)\phantom{\rule{thickmathspace}{0ex}}-\phantom{\rule{thickmathspace}{0ex}}B-C$
${e}^{x}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\mathrm{exp}\left[W\left(A{e}^{B+C}\right)-\phantom{\rule{thickmathspace}{0ex}}B-C\right]$
$q\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\mathrm{exp}\left[W\left(A{e}^{B+C}\right)\phantom{\rule{thickmathspace}{0ex}}-\phantom{\rule{thickmathspace}{0ex}}B-C\right]$
$q\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\mathrm{exp}\left[W\left(\left(\mathrm{ln}n\right){e}^{\mathrm{ln}b+\mathrm{ln}\mathrm{ln}n}\right)\phantom{\rule{thickmathspace}{0ex}}-\phantom{\rule{thickmathspace}{0ex}}\mathrm{ln}b-\mathrm{ln}\mathrm{ln}n\right]$
(a few minutes later) Using ${e}^{\mathrm{ln}b+\mathrm{ln}\mathrm{ln}n}\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}{e}^{\mathrm{ln}b}\cdot {e}^{\mathrm{ln}\mathrm{ln}n}\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}b\mathrm{ln}n,$ we get
$q\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\mathrm{exp}\left[W\left(b{\left(\mathrm{ln}n\right)}^{2}\right)\phantom{\rule{thickmathspace}{0ex}}-\phantom{\rule{thickmathspace}{0ex}}\mathrm{ln}b-\mathrm{ln}\mathrm{ln}n\right]$
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