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Solve the following indices learning, answer in index form.

Mathematics

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4 parts
a

35×713^5 \times 7^1

b

2102^{10}

c

x=719x = \frac{7}{19}

d

x=0x = 0

Step-by-step solution

19 steps

Huu hapa ni utatuzi wa swali la 1, sehemu a, b, c, na d:

1. Tatua viashiria vifuatavyo ukiacha jibu lako katika mfumo wa viashiria:

Part a· 3 steps

243×497\frac{243 \times 49}{7}
  1. Andika nambari zote kama nguvu za nambari kuu.

    243=3×3×3×3×3=35243 = 3 \times 3 \times 3 \times 3 \times 3 = 3^5
    49=7×7=7249 = 7 \times 7 = 7^2
    7=717 = 7^1

  2. Badilisha nambari hizi kwenye usemi.

    35×7271\frac{3^5 \times 7^2}{7^1}
  3. Tumia sheria ya viashiria aman=am−n\frac{a^m}{a^n} = a^{m-n}.

    35×72−1=35×713^5 \times 7^{2-1} = 3^5 \times 7^1

    Jibu katika mfumo wa viashiria ni:

    35×71\boxed{3^5 \times 7^1}

Part b

2102^{10}

1024×25664×4\frac{1024 \times 256}{64 \times 4}
  1. Andika nambari zote kama nguvu za 2.

    1024=2101024 = 2^{10}
    256=28256 = 2^8
    64=2664 = 2^6
    4=224 = 2^2

  2. Badilisha nambari hizi kwenye usemi.

    210×2826×22\frac{2^{10} \times 2^8}{2^6 \times 2^2}
  3. Tumia sheria ya viashiria am×an=am+na^m \times a^n = a^{m+n} kwa namba ya juu na namba ya chini.

    Namba ya juu: 210×28=210+8=2182^{10} \times 2^8 = 2^{10+8} = 2^{18}
    Namba ya chini: 26×22=26+2=282^6 \times 2^2 = 2^{6+2} = 2^8
    Usemi unakuwa:

    21828\frac{2^{18}}{2^8}
  4. Tumia sheria ya viashiria aman=am−n\frac{a^m}{a^n} = a^{m-n}.

    218−8=2102^{18-8} = 2^{10}

    Jibu katika mfumo wa viashiria ni:

    210\boxed{2^{10}}

Part c

x=719x = \frac{7}{19}

(278)x+7=(94)30x\left(\frac{27}{8}\right)^{x+7} = \left(\frac{9}{4}\right)^{30x}
  1. Andika besi kama nguvu za besi ya kawaida.

    278=3323=(32)3\frac{27}{8} = \frac{3^3}{2^3} = \left(\frac{3}{2}\right)^3
    94=3222=(32)2\frac{9}{4} = \frac{3^2}{2^2} = \left(\frac{3}{2}\right)^2

  2. Badilisha hizi kwenye mlinganyo.

    ((32)3)x+7=((32)2)30x\left(\left(\frac{3}{2}\right)^3\right)^{x+7} = \left(\left(\frac{3}{2}\right)^2\right)^{30x}
  3. Tumia sheria ya viashiria (am)n=amn(a^m)^n = a^{mn}.

    (32)3(x+7)=(32)2(30x)\left(\frac{3}{2}\right)^{3(x+7)} = \left(\frac{3}{2}\right)^{2(30x)} (32)3x+21=(32)60x\left(\frac{3}{2}\right)^{3x+21} = \left(\frac{3}{2}\right)^{60x}
  4. Kwa kuwa besi ni sawa, viashiria lazima viwe sawa.

    3x+21=60x3x + 21 = 60x

  5. Tatua kwa xx.

    21=60x−3x21 = 60x - 3x
    21=57x21 = 57x
    x=2157x = \frac{21}{57}

  6. Rahisisha sehemu.

    x=21÷357÷3=719x = \frac{21 \div 3}{57 \div 3} = \frac{7}{19}
    Jibu ni:

    x=719\boxed{x = \frac{7}{19}}

Part d

x=0x = 0

(625)−x÷(3125)2x=15625x(625)^{-x} \div (3125)^{2x} = 15625^x
  1. Andika nambari zote kama nguvu za 5.

    625=54625 = 5^4
    3125=553125 = 5^5
    15625=5615625 = 5^6

  2. Badilisha hizi kwenye mlinganyo.

    (54)−x÷(55)2x=(56)x(5^4)^{-x} \div (5^5)^{2x} = (5^6)^x
  3. Tumia sheria ya viashiria (am)n=amn(a^m)^n = a^{mn}.

    5−4x÷510x=56x5^{-4x} \div 5^{10x} = 5^{6x}
  4. Tumia sheria ya viashiria aman=am−n\frac{a^m}{a^n} = a^{m-n} kwa upande wa kushoto.

    5−4x−10x=56x5^{-4x - 10x} = 5^{6x} 5−14x=56x5^{-14x} = 5^{6x}
  5. Kwa kuwa besi ni sawa, viashiria lazima viwe sawa.

    −14x=6x-14x = 6x

  6. Tatua kwa xx.

    0=6x+14x0 = 6x + 14x
    0=20x0 = 20x
    x=020x = \frac{0}{20}
    x=0x = 0
    Jibu ni:

    x=0\boxed{x = 0}
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