Question

75/1+15e^-0.4x=25 solve for x

Answer 1

Answer:

$\frac{75}{1}+15e^-\cdot \:0.4x=25\quad :\quad x=-\frac{25}{3};\quad \ne \mathrm{True},\:e\ne \:0\quad \left(\mathrm{Decimal}:\quad x=-8.33333\dots \right)$

Step-by-step explanation:

$\frac{75}{1}+15e^-\cdot \:0.4x=25$

$\mathrm{Apply\:rule}\:\frac{a}{1}=a$

$\frac{75}{1}=75$

$75+15\cdot \:0.4e^-x=25$

$\mathrm{Multiply\:the\:numbers:}\:15\cdot \:0.4=6$

$75+6e^-x=25$

$\mathrm{Subtract\:}75\mathrm{\:from\:both\:sides}$

$75+6e^-x-75=25-75$

$\mathrm{Simplify}$

$6e^-x=-50$

$\mathrm{Divide\:both\:sides\:by\:}6e^-;\quad \ne \mathrm{True}$

$\frac{6e^-x}{6e^-}=\frac{-50}{6e^-};\quad \ne \mathrm{True}$

$\mathrm{Simplify}$

$e^{ }x=-\frac{25}{3e^-};\quad \ne \mathrm{True}$

$\mathrm{Divide\:both\:sides\:by\:}e;\quad \ne \mathrm{True},\:e\ne \:0$

$\frac{e^{ }x}{e}=\frac{-\frac{25}{3e^-}}{e};\quad \ne \mathrm{True},\:e\ne \:0$

$\mathrm{Simplify}$

$\frac{x}{e}=-\frac{25}{3e^1};\quad \ne \mathrm{True},\:e\ne \:0$

$\mathrm{Multiply\:both\:sides\:by\:}e$

$\frac{xe}{e}=\left(-\frac{25}{3e^1}\right)e;\quad \ne \mathrm{True},\:e\ne \:0$

$\mathrm{Simplify}$

$x=-\frac{25}{3};\quad \ne \mathrm{True},\:e\ne \:0$