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Question Number 198124 by stevoh last updated on 10/Oct/23
solve for x log100+log(2+x)=10
$${solve}\:{for}\:{x}\:{log}\mathrm{100}+{log}\left(\mathrm{2}+{x}\right)=\mathrm{10} \\ $$
Answered by a.lgnaoui last updated on 10/Oct/23
xlog100+log(2+x)=10 x=10−log(2+x) log100=2 x=((10−log(2+x))/2) (x+2)=5−((log(2+x))/2)+2 (log100=2) x+2=z z=7+−((log z)/2) ((2z+logz)/2)=7 ⇒ log2z=14+log2−2z log2z=14+log2−2z par fonction Wolfram Alpha log2z en fonction de 2z z=6,59053952 ⇒x=4,59053952
$$\:\boldsymbol{\mathrm{xlog}}\mathrm{100}+\boldsymbol{\mathrm{log}}\left(\mathrm{2}+\boldsymbol{\mathrm{x}}\right)=\mathrm{10} \\ $$$$ \\ $$$$\:\:\:\:\mathrm{x}=\mathrm{10}−\mathrm{log}\left(\mathrm{2}+\mathrm{x}\right)\:\:\:\:\:\mathrm{log100}=\mathrm{2} \\ $$$$\:\:\mathrm{x}=\frac{\mathrm{10}−\mathrm{log}\left(\mathrm{2}+\mathrm{x}\right)}{\mathrm{2}} \\ $$$$\left(\mathrm{x}+\mathrm{2}\right)=\mathrm{5}−\frac{\mathrm{log}\left(\mathrm{2}+\mathrm{x}\right)}{\mathrm{2}}+\mathrm{2}\:\:\:\left(\mathrm{log100}=\mathrm{2}\right) \\ $$$$\mathrm{x}+\mathrm{2}=\mathrm{z}\:\:\: \\ $$$$\:\mathrm{z}=\mathrm{7}+−\frac{\mathrm{log}\:\boldsymbol{\mathrm{z}}}{\mathrm{2}} \\ $$$$\:\:\:\frac{\mathrm{2}\boldsymbol{\mathrm{z}}+\boldsymbol{\mathrm{logz}}}{\mathrm{2}}=\mathrm{7}\:\:\:\Rightarrow\:\:\boldsymbol{\mathrm{log}}\mathrm{2}\boldsymbol{\mathrm{z}}=\mathrm{14}+\boldsymbol{\mathrm{log}}\mathrm{2}−\mathrm{2}\boldsymbol{\mathrm{z}} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{log}}\mathrm{2}\boldsymbol{\mathrm{z}}=\mathrm{14}+\boldsymbol{\mathrm{log}}\mathrm{2}−\mathrm{2}\boldsymbol{\mathrm{z}} \\ $$$$ \\ $$$$\:\:\boldsymbol{\mathrm{par}}\:\boldsymbol{\mathrm{fonction}}\:\boldsymbol{\mathrm{Wolfram}}\:\boldsymbol{\mathrm{Alpha}} \\ $$$$\:\:\:\boldsymbol{\mathrm{log}}\mathrm{2}\boldsymbol{\mathrm{z}}\:\:\boldsymbol{\mathrm{en}}\:\boldsymbol{\mathrm{fonction}}\:\boldsymbol{\mathrm{de}}\:\mathrm{2}\boldsymbol{\mathrm{z}}\:\: \\ $$$$\:\:\:\:\:\:\mathrm{z}=\mathrm{6},\mathrm{59053952} \\ $$$$ \\ $$$$\:\:\Rightarrow\boldsymbol{\mathrm{x}}=\mathrm{4},\mathrm{59053952} \\ $$$$\: \\ $$
Answered by Frix last updated on 11/Oct/23
b^x =ax+c ⇒ x=−(c/a)−(1/(ln b))W (−((ln b)/(b^(c/a) a))) xlog 100 +log (x+2) =10 If log x =ln x 100^x (x+2)=e^(10) ((1/(100)))^x =e^(−10) x+2e^(−10) x=−2+(1/(2ln 10))W (20000e^(10) ln 10) x≈1.87721260171 If log x =log_(10) x 100^x (x+2)=10^(10) ((1/(100)))^x =10^(−10) x+2×10^(−10) x=−2+(1/(2ln 10))W (2×10^(14) ln 10) x≈4.59053951583
$${b}^{{x}} ={ax}+{c}\:\Rightarrow\:{x}=−\frac{{c}}{{a}}−\frac{\mathrm{1}}{\mathrm{ln}\:{b}}{W}\:\left(−\frac{\mathrm{ln}\:{b}}{{b}^{{c}/{a}} {a}}\right) \\ $$$$ \\ $$$${x}\mathrm{log}\:\mathrm{100}\:+\mathrm{log}\:\left({x}+\mathrm{2}\right)\:=\mathrm{10} \\ $$$$ \\ $$$$\mathrm{If}\:\mathrm{log}\:{x}\:=\mathrm{ln}\:{x} \\ $$$$\mathrm{100}^{{x}} \left({x}+\mathrm{2}\right)=\mathrm{e}^{\mathrm{10}} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{100}}\right)^{{x}} =\mathrm{e}^{−\mathrm{10}} {x}+\mathrm{2e}^{−\mathrm{10}} \\ $$$${x}=−\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2ln}\:\mathrm{10}}{W}\:\left(\mathrm{20000e}^{\mathrm{10}} \mathrm{ln}\:\mathrm{10}\right) \\ $$$${x}\approx\mathrm{1}.\mathrm{87721260171} \\ $$$$ \\ $$$$\mathrm{If}\:\mathrm{log}\:{x}\:=\mathrm{log}_{\mathrm{10}} \:{x} \\ $$$$\mathrm{100}^{{x}} \left({x}+\mathrm{2}\right)=\mathrm{10}^{\mathrm{10}} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{100}}\right)^{{x}} =\mathrm{10}^{−\mathrm{10}} {x}+\mathrm{2}×\mathrm{10}^{−\mathrm{10}} \\ $$$${x}=−\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2ln}\:\mathrm{10}}{W}\:\left(\mathrm{2}×\mathrm{10}^{\mathrm{14}} \mathrm{ln}\:\mathrm{10}\right) \\ $$$${x}\approx\mathrm{4}.\mathrm{59053951583} \\ $$

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