How do you solve $9 e^{1.4 p - 10} - 10 = 17$ $9e^(1.4p-10)-10=17$ ?

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Answer 1 · 2018-06-14T12:36:20

$p = \frac{ln (3) + 10}{1.4}$ $p = \frac{ln(3)+10}{1.4}$

Add $10$ $10$ to both sides:

$9 e^{1.4 p - 10} = 27$ $9e^{1.4p-10}=27$

divide both sides by $9$ $9$ :

$e^{1.4 p - 10} = 3$ $e^{1.4p-10}=3$

consider the natural logarithm of both sides, leveraging the fact that $ln (e^{x}) = x$ $ln(e^x) = x$

$1.4 p - 10 = ln (3)$ $1.4p-10 = ln(3)$

Add $10$ $10$ to both sides:

$1.4 p = ln (3) + 10$ $1.4p = ln(3)+10$

Divide both sides by $1.4$ $1.4$ :

$p = \frac{ln (3) + 10}{1.4}$ $p = \frac{ln(3)+10}{1.4}$

How do you solve 9e^(1.4p-10)-10=179e1.4p−10−10=179e^(1.4p-10)-10=17?