How do you solve $53^{x + 1} = 65.4$ $53^(x + 1) = 65.4$ ?

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How do you solve $53^{x + 1} = 65.4$ $53^(x + 1) = 65.4$ ?

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Answer 1 · 2016-04-22T06:15:10

$x \approx 0.053$ $xapprox0.053$

Explanation:

First the the $log$ $log$ of both sides:
$53^{x + 1} = 65.4$ $53^(x+1)=65.4$
${log 53}^{x + 1} = log 65.4$ $log53^(x+1)=log65.4$
Then because of the rule ${log a}^{b} = b log a$ $loga^b=bloga$ , we can simplify and solve:
$(x + 1) log 53 = log 65.4$ $(x+1)log53=log65.4$
$x log 53 + log 53 = log 65.4$ $xlog53+log53=log65.4$
$x log 53 = log 65.4 - log 53$ $xlog53=log65.4-log53$
$x = \frac{log 65.4 - log 53}{log 53}$ $x=(log65.4-log53)/log53$
And if you type this into your calculator you get:
$x \approx 0.053$ $xapprox0.053$

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How do you solve $53^{x + 1} = 65.4$ $53^(x + 1) = 65.4$ ?

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Explanation:

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