Jawaban:
[tex]\begin{aligned}\sf {9}^{x + 1} &= \sf {3}^{5x + 7} \\ \sf ( {3}^{2} ) {}^{x + 1} &= \sf {3}^{5x + 7} \\ \sf {3}^{2x + 2} &= \sf {3}^{5x + 7} \\ \sf \cancel{3} {}^{2x + 2} &= \sf \cancel{3} {}^{5x + 7} \\ \sf 2x + 2 &= \sf 5x + 7 \\ \sf 2x - 5x&= \sf 7 - 2 \\ \sf - 3x&= \sf 5 \\ \sf x&= \sf - \frac{5}{3} \\ \sf x&= \sf \red{ - 1 \frac{2}{3} } \end{aligned}[/tex]
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Jawaban:
Penyelesaian :
[tex]\begin{aligned}\sf {9}^{x + 1} &= \sf {3}^{5x + 7} \\ \sf ( {3}^{2} ) {}^{x + 1} &= \sf {3}^{5x + 7} \\ \sf {3}^{2x + 2} &= \sf {3}^{5x + 7} \\ \sf \cancel{3} {}^{2x + 2} &= \sf \cancel{3} {}^{5x + 7} \\ \sf 2x + 2 &= \sf 5x + 7 \\ \sf 2x - 5x&= \sf 7 - 2 \\ \sf - 3x&= \sf 5 \\ \sf x&= \sf - \frac{5}{3} \\ \sf x&= \sf \red{ - 1 \frac{2}{3} } \end{aligned}[/tex]
'조슈아' (Svt)