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Mkey [24]
3 years ago
15

7D%20" id="TexFormula1" title=" {3}^{x + 2} - {3}^{x} = \frac{8}{9} " alt=" {3}^{x + 2} - {3}^{x} = \frac{8}{9} " align="absmiddle" class="latex-formula">
Mathematics
1 answer:
ser-zykov [4K]3 years ago
6 0

3^{x + 2} - 3^{x} = 3^{x}* 3^{2} - 3^{x} = 9(3^{x}) - 1(3^{x}) = 8(3^{x})

3^{x + 2} - 3^{x} = \frac{8}{9}

⇒ 8(3^{x}) = \frac{8}{9}

⇒ 3^{x} = \frac{1}{9}   <em>multiplied both sides by 8</em>

⇒3^{x} = \frac{1}{3^{2} }

⇒ 3^{x} = 3⁻²

⇒ x = -2

Answer: x = -2

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\rm x=\frac{46}{3}

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Step-by-step explanation:

\huge\begin{array} {|c|} \hline\left. \begin{cases} {   \rm\frac{ 1  }{ 4  }  x+ \frac{ 4  }{ 5  }  y =  3  } \\ {   \rm\frac{ 5  }{ 16  }  x- \frac{ 1  }{ 5  }  y=5  }  \end{cases} \right. \\  \hline \end{array}

\\ \Large \overline{ \rm -   \: Steps \:  using  \: substitution \: - }

\rm \frac{1}{4}x+\frac{4}{5}y=3

\rm \frac{1}{4}x=-\frac{4}{5}y+3

\rm x=4\left(-\frac{4}{5}y+3\right)

\rm x=-\frac{16}{5}y+12

┈

\rm \frac{5}{16}\left(-\frac{16}{5}y+12\right)-\frac{1}{5}y=5

\rm -y+\frac{15}{4}-\frac{1}{5}y=5

\rm -\frac{6}{5}y+\frac{15}{4}=5

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\bold{ y=-\frac{25}{24}}

┈

\rm x=-\frac{16}{5}\left(-\frac{25}{24}\right)+12

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\bold{ x=\frac{46}{3}}

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