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Log2(5x − 3) − 2 = log2(x+1)

Mathematics

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Answer

x=7x = 7

Step-by-step solution

6 steps

Hapa kuna hatua za kutatua mlinganyo wa logariti:

Mlinganyo uliotolewa ni:

log⁡2(5x−3)−2=log⁡2(x+1)\log_2 (5x - 3) - 2 = \log_2 (x + 1)
  1. Panga upya mlinganyo ili kuweka maneno ya logariti upande mmoja.

    log⁡2(5x−3)−log⁡2(x+1)=2\log_2 (5x - 3) - \log_2 (x + 1) = 2
  2. Badilisha namba 2 kuwa logariti yenye base 2.

    Tunajua kuwa 2=log⁡2(22)=log⁡242 = \log_2 (2^2) = \log_2 4.

    log⁡2(5x−3)−log⁡2(x+1)=log⁡24\log_2 (5x - 3) - \log_2 (x + 1) = \log_2 4
  3. Tumia sheria ya logariti log⁡bA−log⁡bB=log⁡b(AB)\log_b A - \log_b B = \log_b \left(\frac{A}{B}\right) kuunganisha maneno ya logariti upande wa kushoto.

    log⁡2(5x−3x+1)=log⁡24\log_2 \left(\frac{5x - 3}{x + 1}\right) = \log_2 4
  4. Kwa kuwa base za logariti ni sawa, tunaweza kulinganisha hoja (arguments) zao.

    5x−3x+1=4\frac{5x - 3}{x + 1} = 4
  5. Tatua mlinganyo kwa xx.

    Anza kwa kuzidisha pande zote mbili kwa (x+1)(x + 1).

    5x−3=4(x+1)5x - 3 = 4(x + 1) 5x−3=4x+45x - 3 = 4x + 4

    Toa 4x4x kutoka pande zote mbili:

    x−3=4x - 3 = 4

    Ongeza 3 kwa pande zote mbili:

    x=7x = 7
  6. Hakikisha suluhisho ni halali kwa kuangalia masharti ya logariti (hoja lazima ziwe chanya).

    Kwa 5x−35x - 3: 5(7)−3=35−3=32>05(7) - 3 = 35 - 3 = 32 > 0 (Halali)
    Kwa x+1x + 1: 7+1=8>07 + 1 = 8 > 0 (Halali)
    Kwa kuwa masharti yote yametimizwa, x=7x = 7 ni suluhisho sahihi.

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