limbusamikshya123 limbusamikshya123
  • 14-09-2020
  • Mathematics
contestada

Can you please solve the following equation?

Can you please solve the following equation class=

Respuesta :

surjithayer10 surjithayer10
  • 14-09-2020

Answer:

x=3.903

Step-by-step explanation:

[tex]2^{x-3}.5^{x-1}=200\\2^x.2^{-3}.5^x.5^{-1}=200\\\frac{2^x}{2^3}.\frac{5^x}{5^1}=200\\\frac{2^x.5^x}{2^3.5}=200\\\frac{(2.5)^x}{8.5}=200\\10^x=200 \times 40\\10^x=8000\\log 10^x=log8000\\x=\frac {log8000}{log10}=log8000=log 8\times log1000=log 8+log 1000=log2^3+log10^3=3log2+3log10=3log2+3 \approx0.903+3 \approx3.903[/tex]

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