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**Jodles** Solve for x $\displaystyle ln\, x=1/5 ln(A-1) - 4 ln\,B +4$, expressed without the log's.

My attempt:

$\displaystyle

ln\, x = \frac{{1}}{5}ln\,(A-1)-4ln\, B+4 \\ $

$\displaystyle ln\, x = ln\,(A-1)^{1/5}-ln\, B^{4}+4 \\$

$\displaystyle ln\, x = ln\,\left(\frac{(A-1)^{1/5}}{B^{4}}\right)+4 \\$

$\displaystyle e^{ln\, x} = e^{ln\,\left(\frac{(A-1)^{1/5}}{B^{4}}\right)+4} \\$

$\displaystyle x = \left(\frac{(A-1)^{1/5}}{B^{4}}\right)\times e^{4}$

Can this be done better; or condensed more?

Thanks!