How do you solve $3^{2 x} = 80$ $3^(2x)=80$ ?

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How do you solve $3^{2 x} = 80$ $3^(2x)=80$ ?

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Answer 1 · 2018-02-15T14:45:08

$x \approx 1.994 to 3 dec. places$ $x~~1.994" to 3 dec. places"$

Explanation:

$note that 3^{2 x} = {(3^{2})}^{x} = 9^{x}$ $"note that "3^(2x)=(3^2)^x=9^x$

$\Rightarrow 9^{x} = 80$ $rArr9^x=80$

$[{log x}^{n} \Leftrightarrow n log x]$ $[logx^nhArrnlogx]$

$take the ln (natural log) of both sides$ $"take the ln (natural log) of both sides"$

${ln 9}^{x} = ln 80$ $ln9^x=ln80$

$\Rightarrow x ln 9 = ln 80$ $rArrxln9=ln80$

$\Rightarrow x = \frac{ln 80}{ln 9} \approx 1.994 to 3 dec. places$ $rArrx=ln80/ln9~~1.994" to 3 dec. places"$

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How do you solve $3^{2 x} = 80$ $3^(2x)=80$ ?

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