How do you solve $log x = \frac{3}{2} log 9 + log 2$ $log x=3/2 log 9+log 2$ ?

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How do you solve $log x = \frac{3}{2} log 9 + log 2$ $log x=3/2 log 9+log 2$ ?

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Answer 1 · 2016-03-03T21:56:18

x = 54

Explanation:

using the following $laws of logarithms$ $color(blue)" laws of logarithms "$

• logx + logy = logxy

• log $x^{n} \Leftrightarrow n log x$ $x^n hArr nlogx$

and if ${log}_{b} x = {log}_{b} y \Rightarrow x = y$ $log_bx = log_by rArr x = y$

#rArr logx = log9^(3/2) + log2

[now $[9^{\frac{3}{2}} = {(\sqrt{9})}^{3} = 3^{3} = 27]$ $[9^(3/2) = (sqrt9)^3 = 3^3 = 27 ]$

$\Rightarrow log x = log (27 \times 2) = log 54 \Rightarrow x = 54$ $rArr logx = log(27xx2) = log54 rArr x = 54$

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How do you solve $log x = \frac{3}{2} log 9 + log 2$ $log x=3/2 log 9+log 2$ ?

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