How do you solve $3^{n + 3} = 14.5$ $3^(n+3)=14.5$ ?

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How do you solve $3^{n + 3} = 14.5$ $3^(n+3)=14.5$ ?

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Answer 1 · 2018-04-23T17:02:02

$n \approx 2.43115$ $n~~2.43115$

Explanation:

$3^{x + 3} = 14.5$ $3^(x+3)=14.5$

$\Rightarrow 3^{n} \cdot 3^{3} = 14.5$ $rArr 3^n * 3^3=14.5$

$\Rightarrow 3^{n} \cdot 27 = 14.5$ $rArr 3^n * 27 = 14.5$

$\Rightarrow 3^{n} = \frac{14.5}{27}$ $rArr 3^n=14.5/27$

and based on the definition of logs
$\Rightarrow n = {log}_{3} (\frac{14.5}{27})$ $rArr n=log_3 (14.5/27)$

This can be evaluated with a calculator or spreadsheet function
(for my spread sheet this would look like $= log (14.5 / 27, 3)$ $=log(14.5"/"27,3)$ )
to obtain the approximation:
$n \approx 2.4311499$ $n~~2.4311499$

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How do you solve $3^{n + 3} = 14.5$ $3^(n+3)=14.5$ ?

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