How do you solve ${(\frac{1}{7})}^{y - 3} = 343$ $(1/7)^(y-3)=343$ ?

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How do you solve ${(\frac{1}{7})}^{y - 3} = 343$ $(1/7)^(y-3)=343$ ?

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Answer 1 · 2016-11-08T18:15:57

Please see the explanation for steps leading to the answer: $y = 0$ $y =0$

Explanation:

Given:

${(\frac{1}{7})}^{y - 3} = 343$ $(1/7)^(y - 3) = 343$

Take the natural logarithm on both sides:

$ln ({(\frac{1}{7})}^{y - 3}) = ln (343)$ $ln((1/7)^(y - 3)) = ln(343)$

Use the property of logarithms ${log}_{b} (a^{c}) = (c) {log}_{b} (a)$ $log_b(a^c) = (c)log_b(a)$ on the left side:

$(y - 3) ln (\frac{1}{7}) = ln (343)$ $(y - 3)ln(1/7) = ln(343)$

Divide both sides by $ln (\frac{1}{7})$ $ln(1/7)$ :

$y - 3 = \frac{ln (343)}{ln (\frac{1}{7})}$ $y - 3 = ln(343)/ln(1/7)$

Use the property of logarithms ${log}_{b} (\frac{1}{a}) = - {log}_{b} (a)$ $log_b(1/a) = -log_b(a)$ on the $ln (\frac{1}{7})$ $ln(1/7)$ :

$y - 3 = \frac{ln (343)}{- ln (7)}$ $y - 3 = ln(343)/-ln(7)$

Add 3 to both sides:

$y = 3 - \frac{ln (343)}{ln (7)}$ $y = 3 - ln(343)/ln(7)$

$y = 0$ $y = 0$

Answer 2 · 2016-11-08T19:42:55

$y = 0$ $y=0$

Explanation:

${(\frac{1}{7})}^{y - 3} = 343$ $(1/7)^(y-3)=343$

now $7^{3} = 343$ $7^3=343$

& ${(\frac{1}{7})}^{- 1} = 7$ $(1/7)^-1=7$

$:.(1/7)^(y-3)=7^(3-y)$

$7^(y-3)=7^3=343$

$=>y-3=0$

$:.y=0$

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How do you solve ${(\frac{1}{7})}^{y - 3} = 343$ $(1/7)^(y-3)=343$ ?

2 Answers

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