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Question Number 15672 by Mr easymsn last updated on 12/Jun/17
 solve the equation:    {5^x } +5x=140  please show workings.....
$$\:{solve}\:{the}\:{equation}: \\ $$$$\:\:\left\{\mathrm{5}^{{x}} \right\}\:+\mathrm{5}{x}=\mathrm{140} \\ $$$${please}\:{show}\:{workings}….. \\ $$
Answered by mrW1 last updated on 13/Jun/17
5x=140−{5^x }≤140 and >139  ⇒x≤140/5=28  ⇒x>139/5=27.8  for x=28:  {5^(28) }+5×28=0+140=140 ok  for x=27.xxx=27+f:  {5^(27+f) }+5f=5  f=1−(1/5){5^(27) ×5^f }  there must exist solutions for f.    so one solution is certain: x=28,  other solutions x=27+f exist, but  can not be calculated.
$$\mathrm{5x}=\mathrm{140}−\left\{\mathrm{5}^{\mathrm{x}} \right\}\leqslant\mathrm{140}\:\mathrm{and}\:>\mathrm{139} \\ $$$$\Rightarrow\mathrm{x}\leqslant\mathrm{140}/\mathrm{5}=\mathrm{28} \\ $$$$\Rightarrow\mathrm{x}>\mathrm{139}/\mathrm{5}=\mathrm{27}.\mathrm{8} \\ $$$$\mathrm{for}\:\mathrm{x}=\mathrm{28}: \\ $$$$\left\{\mathrm{5}^{\mathrm{28}} \right\}+\mathrm{5}×\mathrm{28}=\mathrm{0}+\mathrm{140}=\mathrm{140}\:\mathrm{ok} \\ $$$$\mathrm{for}\:\mathrm{x}=\mathrm{27}.\mathrm{xxx}=\mathrm{27}+\mathrm{f}: \\ $$$$\left\{\mathrm{5}^{\mathrm{27}+\mathrm{f}} \right\}+\mathrm{5f}=\mathrm{5} \\ $$$$\mathrm{f}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\left\{\mathrm{5}^{\mathrm{27}} ×\mathrm{5}^{\mathrm{f}} \right\} \\ $$$$\mathrm{there}\:\mathrm{must}\:\mathrm{exist}\:\mathrm{solutions}\:\mathrm{for}\:\mathrm{f}. \\ $$$$ \\ $$$$\mathrm{so}\:\mathrm{one}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{certain}:\:\mathrm{x}=\mathrm{28}, \\ $$$$\mathrm{other}\:\mathrm{solutions}\:\mathrm{x}=\mathrm{27}+\mathrm{f}\:\mathrm{exist},\:\mathrm{but} \\ $$$$\mathrm{can}\:\mathrm{not}\:\mathrm{be}\:\mathrm{calculated}. \\ $$

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