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Question Number 13152 by tawa tawa last updated on 15/May/17

$$\mathrm{Solve}:5^{\log \left(\mathrm{x}\right)}+{\mathrm{logx}}^5=25$$

Answered by mrW1 last updated on 16/May/17

$$\log x=t$$$$5^t+5t=25$$$$5^t=5(5-t)$$$$5^{t-5}=\frac{5-t}{625}$$$$e^{(t-5)\ln 5}=\frac{5-t}{625}$$$$(5-t)\ln 5e^{(t-5)\ln 5}=625\times \ln 5$$$$(5-t)\ln 5=W\left(625\times \ln 5\right)$$$$t=5-\frac{W\left(625\times \ln 5\right)}{\ln 5}=\log x$$$$\Rightarrow x={10}^{5-\frac{W\left(625\times \ln 5\right)}{\ln 5}}\approx {10}^{5-\frac{5.254544}{\ln 5}}\approx 54.346$$

Commented by tawa tawa last updated on 15/May/17

$$\mathrm{Wow},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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