Question

jika
[tex] \frac{ {3}^{x} }{ {27}^{x + 2} } = 9 \sqrt{ {3}^{x - 3} } . \sqrt{ {3}^{ x + 5} } [/tex]
maka nilai X yang memenuhi adalah....​

Answer 1

Jawaban:

x = -3

Penjelasan dengan langkah-langkah:

[tex]\begin{aligned}\displaystyle\rm~\frac{ {3}^{x} }{ {27}^{x + 2} } & = \displaystyle\rm~ 9 \sqrt{ {3}^{x - 3} } . \sqrt{ {3}^{ x + 5} } \\ \\ \displaystyle\rm~ \frac{ {3}^{x} }{ { ({3}^{3} )}^{x + 2} } & = \displaystyle\rm~ {3}^{2}. {( {3}^{x - 3}) }^{ \frac{1}{2} }. { ({3}^{x + 5}) }^{ \frac{1}{2} } \\ \\ \displaystyle\rm~ \frac{ {3}^{x} }{ {3}^{3x + 6} } & = \displaystyle\rm~ {3}^{2}. {3}^{ \frac{x}{2} - \frac{3}{2} }. {3}^{ \frac{x}{2} + \frac{5}{2} } \\ \\ \displaystyle\rm~ {3}^{x - (3x + 6)} & = \displaystyle\rm~ {3}^{2 + \frac{x}{2} - \frac{3}{2} + \frac{x}{2} + \frac{5}{2} } \\ \displaystyle\rm~ {3}^{x - 3x - 6} & = \displaystyle\rm~ {3}^{2 + \frac{x}{2} + \frac{x}{2} - \frac{3}{2} + \frac{5}{2} } \\ \displaystyle\rm~ {3}^{ - 2x - 6} & = \displaystyle\rm~ {3}^{2 + \frac{2x}{2} + \frac{2}{2} } \\ \displaystyle\rm~ {3}^{ - 2x - 6} & = \displaystyle\rm~ {3}^{2 + x + 1} \\ \displaystyle\rm~ {\cancel{3}}^{ - 2x - 6} & = \displaystyle\rm~ {\cancel{3}}^{x + 3} \\ \displaystyle\rm~ - 2x - 6 & = \displaystyle\rm~x + 3 \\ \displaystyle\rm~ - 2x - x & = \displaystyle\rm~3 + 6 \\ \displaystyle\rm~ - 3x & = \displaystyle\rm~ 9 \\ \\ \displaystyle\rm~x & = \displaystyle\rm~ \frac{9}{ - 3} \\ \\ \displaystyle\rm~x & = \displaystyle\rm~ - 3 \end{aligned}[/tex]