Huu hapa utatuzi wa swali la 35:

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Huu hapa utatuzi wa swali la 35:

Tunapewa mlinganyo:

32^{(x-3)} \div 8^{(x+4)} = 64 \div 2^x

Andika nambari zote kama nguvu za 2.
Tunajua kuwa $32 = 2^5$ , $8 = 2^3$ , na $64 = 2^6$ .
Badilisha nambari hizi kwenye mlinganyo:
$(2^5)^{(x-3)} \div (2^3)^{(x+4)} = 2^6 \div 2^x$
Tumia sheria ya viashiria $(a^m)^n = a^{mn}$ .
$2^{5(x-3)} \div 2^{3(x+4)} = 2^6 \div 2^x$ $2^{(5x-15)} \div 2^{(3x+12)} = 2^6 \div 2^x$
Tumia sheria ya viashiria $a^m \div a^n = a^{m-n}$ .
$2^{(5x-15) - (3x+12)} = 2^{(6-x)}$ $2^{(5x-15-3x-12)} = 2^{(6-x)}$ $2^{(2x-27)} = 2^{(6-x)}$
Kwa kuwa besi (base) ni sawa, linganisha viashiria (exponents).
$2x - 27 = 6 - x$
Tatua kwa $x$ .
Ongeza $x$ pande zote mbili:
$2x + x - 27 = 6$ $3x - 27 = 6$
Ongeza 27 pande zote mbili:
$3x = 6 + 27$ $3x = 33$
Gawanya kwa 3:
$x = \frac{33}{3}$ $x = \boxed{\text{11}}$

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