# How do you solve (1/7)^(y-3)=343?

Nov 8, 2016

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

#### Explanation:

Given:

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

Take the natural logarithm on both sides:

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

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

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

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

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

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

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

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

$y = 0$

Nov 8, 2016

$y = 0$

#### Explanation:

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

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

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

$\therefore {\left(\frac{1}{7}\right)}^{y - 3} = {7}^{3 - y}$

${7}^{y - 3} = {7}^{3} = 343$

$\implies y - 3 = 0$

$\therefore y = 0$