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Find the exact solutions, in their simplest form, to the equations (a) 2 ln(2x + 1) - 10 = 0 (b) 3^e * e^x = e^7 - Edexcel - A-Level Maths Pure - Question 4 - 2014 - Paper 5

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Find the exact solutions, in their simplest form, to the equations (a) 2 ln(2x + 1) - 10 = 0 (b) 3^e * e^x = e^7

Worked Solution & Example Answer:Find the exact solutions, in their simplest form, to the equations (a) 2 ln(2x + 1) - 10 = 0 (b) 3^e * e^x = e^7 - Edexcel - A-Level Maths Pure - Question 4 - 2014 - Paper 5

Step 1

(a) 2 ln(2x + 1) - 10 = 0

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Answer

To solve for x, start by isolating the logarithmic term:

  1. Add 10 to both sides:

    2ln⁡(2x+1)=102 \ln(2x + 1) = 10

  2. Divide by 2:

    ln⁡(2x+1)=5\ln(2x + 1) = 5

  3. Exponentiate both sides to eliminate the logarithm:

    2x+1=e52x + 1 = e^5

  4. Subtract 1 from both sides:

    2x=e5−12x = e^5 - 1

  5. Finally, divide by 2:

    x=e5−12x = \frac{e^5 - 1}{2}

Step 2

(b) 3^e * e^x = e^7

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Answer

To find x, we can follow these steps:

  1. Rewriting the equation:

    3e⋅ex=e73^e \cdot e^x = e^7

  2. Divide both sides by 3e3^e:

    ex=e73ee^x = \frac{e^7}{3^e}

  3. Take the natural logarithm of both sides:

    x=7−ln⁡(3e)x = 7 - \ln(3^e)

  4. By using the logarithm property, ln⁡(ab)=bln⁡(a)\ln(a^b) = b\ln(a):

    x=7−eln⁡(3)x = 7 - e \ln(3)

  5. Therefore, the solution is:

    x=7−eln⁡(3)x = 7 - e \ln(3)

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