How do you solve $3 \cdot 18^{n + 8} = 1$ $3*18^(n+8)=1$ ?

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How do you solve $3 \cdot 18^{n + 8} = 1$ $3*18^(n+8)=1$ ?

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Answer 1 · 2016-08-24T20:17:53

$n = - (8 + \frac{1}{{log}_{3} 2 + 2})$ $n = -(8+1/(log_3 2 + 2))$

Explanation:

$3 \cdot 18^{n + 8} = 1$ $3*18^(n+8)=1$ . Applying the $log$ $log$ transformation to both sides

${log}_{e} 3 + (n + 8) {log}_{e} 18 = 0$ $log_e3+(n+8)log_e 18 = 0$ or
$n = - \frac{8 {log}_{e} 18 + {log}_{e} 3}{{log}_{e} 18} = - 8 - \frac{{log}_{e} 3}{{log}_{e} 18}$ $n = -(8log_e 18+log_e 3)/(log_e 18) = -8-log_e3/(log_e 18)$ then

$n = - 8 - \frac{{log}_{e} 3}{{log}_{e} 2 + 2 {log}_{e} 3} = - 8 - \frac{1}{\frac{{log}_{e} 2}{{log}_{e} 3} + 2}$ $n = -8-log_e3/(log_e2+2log_e 3) = -8-1/(log_e2/log_e3+2$

Finally

$n = - (8 + \frac{1}{{log}_{3} 2 + 2})$ $n = -(8+1/(log_3 2 + 2))$

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How do you solve $3 \cdot 18^{n + 8} = 1$ $3*18^(n+8)=1$ ?

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