It is a series and sequences question. I worked it up to here but I'm stuck at this point. $$\left(\dfrac{4}{3}\right)^n = \dfrac{6439}{3072}$$ Solve for $n$. Thanks
2026-05-15 15:56:52.1778860612
Solve for $n$ in $(4/3) ^n = 6439 / 3072 $
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$\left(\dfrac{4}{3}\right)^n=\dfrac{6439}{3072}\\\implies\log \left(\dfrac{4}{3}\right)^n=\log \dfrac{6439}{3072}\\\implies n\cdot \log\dfrac{4}{3}=\log \dfrac{6439}{3072}\\\implies n=\dfrac{\log \dfrac{6439}{3072}}{\log\dfrac{4}{3}}\\\implies n= 2.572438483$