How do you solve $400 {(1 - 0.2)}^{x} = 50$ $400(1-0.2)^x=50$ ?

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How do you solve $400 {(1 - 0.2)}^{x} = 50$ $400(1-0.2)^x=50$ ?

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Answer 1 · 2018-05-09T14:52:27

$x = \frac{ln (8)}{ln (\frac{5}{4})}$ $x=ln(8)/ln(5/4)$

Explanation:

$400 {(1 - 0.2)}^{x} = 50$ $400(1-0.2)^x=50$
$ln (400 {(1 - 0.2)}^{x}) = ln 50$ $ln(400(1-0.2)^x)=ln50$
$ln a b = ln a + ln b$ $lnab=lna+lnb$
$ln 400 + ln ({0.8}^{x}) = ln 50$ $ln400+ln(0.8^x)=ln50$
${ln a}^{b} = b ln a$ $lna^b=blna$
$ln 400 + x ln 0.8 = ln 50$ $ln400+xln0.8=ln50$
$x ln 0.8 = ln (\frac{50}{400})$ $xln0.8=ln(50/400)$
$x ln 0.8 = ln (\frac{1}{8})$ $xln0.8=ln(1/8)$
$cancel(-)xln(5/4)=cancel(-)ln8$
$x=ln(8)/ln(5/4)$
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How do you solve $400 {(1 - 0.2)}^{x} = 50$ $400(1-0.2)^x=50$ ?

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Explanation:

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How do you solve $400 {(1 - 0.2)}^{x} = 50$ $400(1-0.2)^x=50$ ?