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ScanSolve AI Tutor · Accepted Answer

Huu hapa utatuzi wa swali la 33:

Tunapewa mlinganyo:

\log_2 (x+1) - \log_2 (x-1) = 3

Tumia sheria ya logariti $\log_b A - \log_b B = \log_b \left(\frac{A}{B}\right)$ .
$\log_2 \left(\frac{x+1}{x-1}\right) = 3$
Badilisha mlinganyo wa logariti kuwa mlinganyo wa kielelezo (exponential equation).
Kumbuka kwamba $\log_b y = x$ ni sawa na $b^x = y$ .
Hivyo,
$2^3 = \frac{x+1}{x-1}$
Rahisisha na tatua kwa $x$ .
$8 = \frac{x+1}{x-1}$
Zidisha pande zote mbili kwa $(x-1)$ :
$8(x-1) = x+1$ $8x - 8 = x+1$
Toa $x$ kutoka pande zote mbili:
$7x - 8 = 1$
Ongeza 8 pande zote mbili:
$7x = 9$
Gawanya kwa 7:
$x = \frac{9}{7}$
Hakikisha thamani ya $x$ inafanya hoja za logariti kuwa chanya.
Kwa $x = \frac{9}{7}$ :
$x+1 = \frac{9}{7} + 1 = \frac{9+7}{7} = \frac{16}{7} > 0$
$x-1 = \frac{9}{7} - 1 = \frac{9-7}{7} = \frac{2}{7} > 0$
Thamani ya $x$ ni halali.
$x = \boxed{\frac{9}{7}}$

Answer