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Question Number 109626 by qwerty111 last updated on 24/Aug/20

Commented by kaivan.ahmadi last updated on 24/Aug/20

$${log}_{20}2\times 5^3={log}_{20}2+3{log}_{20}5=\frac{1}{{log}_{2}20}+\frac{3}{{log}_{5}20}=$$$$\frac{1}{2+{log}_{2}5}+\frac{3}{1+2{log}_{5}2}=m$$$$letx={log}_{2}5\Rightarrow$$$$\frac{1}{2+x}+\frac{3}{1+\frac{2}{x}}=m\Rightarrow \frac{1}{2+x}+\frac{3x}{2+x}=m\Rightarrow$$$$\frac{1+3x}{x+2}=m\Rightarrow 1+3x=mx+2m\Rightarrow (m-3)x=1-2m\Rightarrow$$$$x=\frac{1-2m}{m-3}$$$$$$

Answered by Rasheed.Sindhi last updated on 24/Aug/20

$$\mathrm{m}=\frac{{\log }_{2}250}{{\log }_{2}20}=\frac{{\log }_2(5^3.2)}{{\log }_2(5.2^2)}$$$$=\frac{{3\log }_{2}5+{\log }_{2}2}{{\log }_{2}5+{2\log }_{2}2}=\frac{{3\log }_{2}5+1}{{\log }_{2}5+2}$$$$\frac{{3\log }_{2}5+1}{{\log }_{2}5+2}=\mathrm{m}$$$${3\log }_{2}5+1={\mathrm{mlog}}_{2}5+2\mathrm{m}$$$${3\log }_{2}5-{\mathrm{mlog}}_{2}5=2\mathrm{m}-1$$$$(3-\mathrm{m}){\log }_{2}5=2\mathrm{m}-1$$$${\log }_{2}5=\frac{2\mathrm{m}-1}{3-\mathrm{m}}=\frac{1-2\mathrm{m}}{\mathrm{m}-3}$$$$$$

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