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Solve for x given that 1/2 log, 9 + log, (5x−4) = 7. The figure below represents part of a school badge. The badge has rotation symmetry of order 4 and 0. Complete the figure to show the badge.

Mathematics

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Answer

1

x=283 au 913x = \frac{28}{3} \text{ au } 9 \frac{1}{3}

Step-by-step solution

7 steps

Hapa kuna hatua za kutatua maswali uliyotoa:

Part 1· 7 steps

Tatua kwa xx kutokana na 12log⁡29+log⁡2(5x−4)=7\frac{1}{2} \log_2 9 + \log_2 (5x-4) = 7.

  1. Rahisisha neno la kwanza kwa kutumia sheria ya logariti log⁡b(Mk)=klog⁡bM\log_b (M^k) = k \log_b M.

    12log⁡29=log⁡2(91/2)=log⁡29=log⁡23\frac{1}{2} \log_2 9 = \log_2 (9^{1/2}) = \log_2 \sqrt{9} = \log_2 3
  2. Badilisha neno lililorahisishwa kwenye mlinganyo wa asili.

    log⁡23+log⁡2(5x−4)=7\log_2 3 + \log_2 (5x-4) = 7
  3. Tumia sheria ya logariti log⁡bM+log⁡bN=log⁡b(MN)\log_b M + \log_b N = \log_b (MN) kuunganisha maneno ya logariti.

    log⁡2(3(5x−4))=7\log_2 (3(5x-4)) = 7 log⁡2(15x−12)=7\log_2 (15x-12) = 7
  4. Badilisha mlinganyo wa logariti kuwa mlinganyo wa kielelezo.

    Ikiwa log⁡bM=N\log_b M = N, basi bN=Mb^N = M.

    15x−12=2715x-12 = 2^7
  5. Kokotoa thamani ya 272^7.

    15x−12=12815x-12 = 128
  6. Tatua mlinganyo wa mstari kwa xx.

    15x=128+1215x = 128 + 12 15x=14015x = 140 x=14015x = \frac{140}{15}
  7. Rahisisha sehemu.

    x=283x = \frac{28}{3}

    Unaweza pia kuandika jibu kama nambari mchanganyiko.

    x=913x = 9 \frac{1}{3}

    Jibu la mwisho ni:

    x=283 au 913\boxed{x = \frac{28}{3} \text{ au } 9 \frac{1}{3}}

Part 2

Kamilisha umbo kuonyesha beji.

Umbo lina ulinganifu wa mzunguko wa mpangilio wa 4 kuhusu kituo O. Hii inamaanisha kuwa beji nzima itaonekana sawa baada ya kuzungushwa kwa 90∘90^\circ, 180∘180^\circ, na 270∘270^\circ kuzunguka O. Sehemu uliyopewa ni moja ya sehemu nne zinazofanana zinazounda beji.

Ili kukamilisha umbo:

  • Chora nakala tatu zaidi za pembetatu uliyopewa.
  • Kila nakala inapaswa kupatikana kwa kuzungusha pembetatu ya asili kwa 90∘90^\circ, 180∘180^\circ, na 270∘270^\circ mtawalia, kuzunguka kituo O.
  • Pembetatu zote nne zitakutana kwenye kituo O. Kila pembetatu itakuwa na pande zenye urefu wa 3 cm na 4 cm zinazotoka O, na upande wa 2 cm unaounganisha ncha za pande hizo mbili.
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