Solve the value of x. i) log₂(x−1) + log₂x = 6; ii) log₃(x+1) + log₃(x+2) = 3.

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Hapa kuna hatua za kutatua milinganyo ya logariti: i) log₂ (x−1) + log₂ x = 6; Step 1: Tumia sheria ya logariti log_b M + log_b N = log_b (M · N) kuunganisha maneno. log₂ ((x−1)x) = 6; log₂ (x² − x) = 6; Step 2: Badilisha mlinganyo wa logariti kuwa mlinganyo wa kielelezo kwa kutumia log_b A = C ⇒ b^C = A . 2⁶ = x² − x; 64 = x² − x; Step 3: Panga upya mlinganyo kuwa fomu ya quadratic ax² + bx + c = 0. x² − x − 64 = 0; Step 4: Tumia fomula ya quadratic x = (−b ± √(b² − 4ac))/(2a) kutatua kwa x. Hapa a=1, b=−1, c=−64. x = (−(−1) ± √((−1)² − 4(1)(−64)))/(2(1)); x = (1 ± √(1 + 256))/(2); x = (1 ± √(257))/(2) Step 5: Angalia masharti ya logariti. Kwa log₂ (x−1) na log₂ x kufafanuliwa, tunahitaji x−1 > 0 ⇒ x > 1 na x > 0. Hivyo, x lazima iwe kubwa kuliko 1. x₁ = (1 + √(257))/(2) ≈ (1 + 16.03)/(2) ≈ 8.515; x₂ = (1 − √(257))/(2) ≈ (1 − 16.03)/(2) ≈ −7.515 Thamani ya x₂ si kubwa kuliko 1, kwa hivyo si suluhisho halali. Suluhisho la kwanza ni: x = (1 + √(257))/(2); ii) log₃ (x+1) + log₃ (x+2) = 3; Step 1: Tumia sheria ya logariti log_b M + log_b N = log_b (M · N) kuunganisha maneno. log₃ ((x+1)(x+2)) = 3; log₃ (x² + 2x + x + 2) = 3; log₃ (x² + 3x + 2) = 3; Step 2: Badilisha mlinganyo wa logariti kuwa mlinganyo wa kielelezo kwa kutumia log_b A = C ⇒ b^C = A . 3³ = x² + 3x + 2; 27 = x² + 3x + 2; Step 3: Panga upya mlinganyo kuwa fomu ya quadratic ax² + bx + c = 0. x² + 3x + 2 − 27 = 0; x² + 3x − 25 = 0; Step 4: Tumia fomula ya quadratic x = (−b ± √(b² − 4ac))/(2a) kutatua kwa x. Hapa a=1, b=3, c=−25. x = (−3 ± √(3² − 4(1)(−25)))/(2(1)); x = (−3 ± √(9 + 100))/(2); x = (−3 ± √(109))/(2) Step 5: Angalia masharti ya logariti. Kwa log₃ (x+1) na log₃ (x+2) kufafanuliwa, tunahitaji x+1 > 0 ⇒ x > −1 na x+2 > 0 ⇒ x > −2. Hivyo, x lazima iwe kubwa kuliko −1. x₁ = (−3 + √(109))/(2) ≈ (−3 + 10.44)/(2) ≈ 3.72; x₂ = (−3 − √(109))/(2) ≈ (−3 − 10.44)/(2) ≈ −6.72 Thamani ya x₂ si kubwa kuliko −1, kwa hivyo si suluhisho halali. Suluhisho la pili ni: x = (−3 + √(109))/(2)

Solve the value of x. i) log₂(x−1) + log₂x = 6; ii) log₃(x+1) + log₃(x+2) = 3.

Answers

Step-by-step solution

Part i· 5 steps

Tumia sheria ya logariti $\log_b M + \log_b N = \log_b (M \cdot N)$ kuunganisha maneno.

Badilisha mlinganyo wa logariti kuwa mlinganyo wa kielelezo kwa kutumia $\log_b A = C \implies b^C = A$ .

Panga upya mlinganyo kuwa fomu ya quadratic $ax^2 + bx + c = 0$ .

Tumia fomula ya quadratic $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ kutatua kwa $x$ .

Angalia masharti ya logariti.

Part ii· 5 steps
$x = \frac{-3 + \sqrt{109}}{2}$

Tumia sheria ya logariti $\log_b M + \log_b N = \log_b (M \cdot N)$ kuunganisha maneno.

Badilisha mlinganyo wa logariti kuwa mlinganyo wa kielelezo kwa kutumia $\log_b A = C \implies b^C = A$ .

Panga upya mlinganyo kuwa fomu ya quadratic $ax^2 + bx + c = 0$ .

Tumia fomula ya quadratic $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ kutatua kwa $x$ .

Angalia masharti ya logariti.

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Solve the value of x. i) log₂(x−1) + log₂x = 6; ii) log₃(x+1) + log₃(x+2) = 3.

Answers

Step-by-step solution

Part i· 5 steps

Tumia sheria ya logariti log⁡bM+log⁡bN=log⁡b(M⋅N)\log_b M + \log_b N = \log_b (M \cdot N)logb​M+logb​N=logb​(M⋅N) kuunganisha maneno.

Badilisha mlinganyo wa logariti kuwa mlinganyo wa kielelezo kwa kutumia log⁡bA=C ⟹ bC=A\log_b A = C \implies b^C = Alogb​A=C⟹bC=A.

Panga upya mlinganyo kuwa fomu ya quadratic ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0.

Tumia fomula ya quadratic x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}x=2a−b±b2−4ac​​ kutatua kwa xxx.

Angalia masharti ya logariti.

Part ii· 5 stepsx=−3+1092x = \frac{-3 + \sqrt{109}}{2}x=2−3+109​​

Tumia sheria ya logariti log⁡bM+log⁡bN=log⁡b(M⋅N)\log_b M + \log_b N = \log_b (M \cdot N)logb​M+logb​N=logb​(M⋅N) kuunganisha maneno.

Badilisha mlinganyo wa logariti kuwa mlinganyo wa kielelezo kwa kutumia log⁡bA=C ⟹ bC=A\log_b A = C \implies b^C = Alogb​A=C⟹bC=A.

Panga upya mlinganyo kuwa fomu ya quadratic ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0.

Tumia fomula ya quadratic x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}x=2a−b±b2−4ac​​ kutatua kwa xxx.

Angalia masharti ya logariti.

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Still stuck on this one?

More Mathematics Questions

Tumia sheria ya logariti $\log_b M + \log_b N = \log_b (M \cdot N)$ kuunganisha maneno.

Badilisha mlinganyo wa logariti kuwa mlinganyo wa kielelezo kwa kutumia $\log_b A = C \implies b^C = A$ .

Panga upya mlinganyo kuwa fomu ya quadratic $ax^2 + bx + c = 0$ .

Tumia fomula ya quadratic $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ kutatua kwa $x$ .

Part ii· 5 steps
$x = \frac{-3 + \sqrt{109}}{2}$

Tumia sheria ya logariti $\log_b M + \log_b N = \log_b (M \cdot N)$ kuunganisha maneno.

Badilisha mlinganyo wa logariti kuwa mlinganyo wa kielelezo kwa kutumia $\log_b A = C \implies b^C = A$ .

Panga upya mlinganyo kuwa fomu ya quadratic $ax^2 + bx + c = 0$ .

Tumia fomula ya quadratic $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ kutatua kwa $x$ .