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Question Number 198124 by stevoh last updated on 10/Oct/23

$$solveforxlog100+log(2+x)=10$$

Answered by a.lgnaoui last updated on 10/Oct/23

$$\mathbf{xlog}100+\mathbf{log}(2+\mathbf{x})=10$$$$$$$$\mathrm{x}=10-\log (2+\mathrm{x})\log 100=2$$$$\mathrm{x}=\frac{10-\log (2+\mathrm{x})}{2}$$$$(\mathrm{x}+2)=5-\frac{\log (2+\mathrm{x})}{2}+2(\log 100=2)$$$$\mathrm{x}+2=\mathrm{z}$$$$\mathrm{z}=7+-\frac{\log \mathbf{z}}{2}$$$$\frac{2\mathbf{z}+\mathbf{logz}}{2}=7\Rightarrow \mathbf{log}2\mathbf{z}=14+\mathbf{log}2-2\mathbf{z}$$$$$$$$\mathbf{log}2\mathbf{z}=14+\mathbf{log}2-2\mathbf{z}$$$$$$$$\mathbf{par}\mathbf{fonction}\mathbf{Wolfram}\mathbf{Alpha}$$$$\mathbf{log}2\mathbf{z}\mathbf{en}\mathbf{fonction}\mathbf{de}2\mathbf{z}$$$$\mathrm{z}=6,59053952$$$$$$$$\Rightarrow \mathbf{x}=4,59053952$$$$$$

Answered by Frix last updated on 11/Oct/23

$$b^x=ax+c\Rightarrow x=-\frac{c}{a}-\frac{1}{\ln b}W(-\frac{\ln b}{b^{c/a}a})$$$$$$$$x\log 100+\log (x+2)=10$$$$$$$$\mathrm{If}\log x=\ln x$$$${100}^x(x+2)={\mathrm{e}}^{10}$$$${\left(\frac{1}{100}\right)}^x={\mathrm{e}}^{-10}x+{2\mathrm{e}}^{-10}$$$$x=-2+\frac{1}{2\ln 10}W\left({20000\mathrm{e}}^{10}\ln 10\right)$$$$x\approx 1.87721260171$$$$$$$$\mathrm{If}\log x={\log }_{10}x$$$${100}^x(x+2)={10}^{10}$$$${\left(\frac{1}{100}\right)}^x={10}^{-10}x+2\times {10}^{-10}$$$$x=-2+\frac{1}{2\ln 10}W\left(2\times {10}^{14}\ln 10\right)$$$$x\approx 4.59053951583$$

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