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Simplify the following Surds: a) Simplify 6/(2sqrt(2)-1). b) Simplify (2sqrt(3)+2)/(2sqrt(3)-2).

Mathematics

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Here are the solutions to the problems:

Q1: Simplify the following Surds.

a) Simplify 622−1\frac{6}{2\sqrt{2}-1}.

  1. Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is 22+12\sqrt{2}+1.

    622−1×22+122+1\frac{6}{2\sqrt{2}-1} \times \frac{2\sqrt{2}+1}{2\sqrt{2}+1}
  2. Multiply the numerators and denominators.

    6(22+1)(22−1)(22+1)\frac{6(2\sqrt{2}+1)}{(2\sqrt{2}-1)(2\sqrt{2}+1)}
  3. Expand the numerator and use the difference of squares formula (a−b)(a+b)=a2−b2(a-b)(a+b)=a^2-b^2 for the denominator.

    122+6(22)2−12\frac{12\sqrt{2}+6}{(2\sqrt{2})^2 - 1^2}
  4. Simplify the denominator.

    122+64×2−1=122+68−1=122+67\frac{12\sqrt{2}+6}{4 \times 2 - 1} = \frac{12\sqrt{2}+6}{8 - 1} = \frac{12\sqrt{2}+6}{7}

    The simplified expression is:

    122+67\boxed{\frac{12\sqrt{2}+6}{7}}

    b) Simplify 23+223−2\frac{2\sqrt{3}+2}{2\sqrt{3}-2}.

  5. Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is 23+22\sqrt{3}+2.

    23+223−2×23+223+2\frac{2\sqrt{3}+2}{2\sqrt{3}-2} \times \frac{2\sqrt{3}+2}{2\sqrt{3}+2}
  6. Multiply the numerators and denominators.

    (23+2)(23+2)(23−2)(23+2)\frac{(2\sqrt{3}+2)(2\sqrt{3}+2)}{(2\sqrt{3}-2)(2\sqrt{3}+2)}
  7. Expand the numerator using (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2 and the denominator using (a−b)(a+b)=a2−b2(a-b)(a+b)=a^2-b^2.

    (23)2+2(23)(2)+22(23)2−22\frac{(2\sqrt{3})^2 + 2(2\sqrt{3})(2) + 2^2}{(2\sqrt{3})^2 - 2^2}
  8. Simplify the terms.

    4×3+83+44×3−4=12+83+412−4\frac{4 \times 3 + 8\sqrt{3} + 4}{4 \times 3 - 4} = \frac{12 + 8\sqrt{3} + 4}{12 - 4}
  9. Combine like terms.

    16+838\frac{16 + 8\sqrt{3}}{8}
  10. Factor out the common term from the numerator and simplify.

    8(2+3)8=2+3\frac{8(2 + \sqrt{3})}{8} = 2 + \sqrt{3}

    The simplified expression is:

    2+3\boxed{2+\sqrt{3}}

    Q2: Simplify the following Logarithms.

    a) Solve log⁡4(2x+3)−3=log⁡4x\log_4 (2x+3) - 3 = \log_4 x.

  11. Rearrange the equation to group the logarithm terms on one side.

    log⁡4(2x+3)−log⁡4x=3\log_4 (2x+3) - \log_4 x = 3
  12. Use the logarithm property log⁡bA−log⁡bB=log⁡b(AB)\log_b A - \log_b B = \log_b \left(\frac{A}{B}\right).

    log⁡4(2x+3x)=3\log_4 \left(\frac{2x+3}{x}\right) = 3
  13. Convert the logarithmic equation to an exponential equation.

    If log⁡bA=C\log_b A = C, then A=bCA = b^C.

    2x+3x=43\frac{2x+3}{x} = 4^3
  14. Calculate 434^3.

    2x+3x=64\frac{2x+3}{x} = 64
  15. Multiply both sides by xx.

    2x+3=64x2x+3 = 64x
  16. Subtract $2x$ from both sides.

    3=64x−2x3 = 64x - 2x 3=62x3 = 62x
  17. Solve for xx.

    x=362x = \frac{3}{62}

    The solution for xx is:

    x=362\boxed{x = \frac{3}{62}}

    b) If log⁡2y=3−log⁡243/2\log_2 y = 3 - \log_2 4^{3/2}, find the value of yy.

  18. Simplify the term 43/24^{3/2}.

    43/2=(4)3=23=84^{3/2} = (\sqrt{4})^3 = 2^3 = 8
  19. Substitute this value back into the equation.

    log⁡2y=3−log⁡28\log_2 y = 3 - \log_2 8
  20. Evaluate log⁡28\log_2 8.

    Since 23=82^3 = 8, log⁡28=3\log_2 8 = 3.

    log⁡2y=3−3\log_2 y = 3 - 3 log⁡2y=0\log_2 y = 0
  21. Convert the logarithmic equation to an exponential equation.

    y=20y = 2^0
  22. Calculate 202^0.

    y=1y = 1

    The value of yy is:

    y=1\boxed{y=1}
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