High School

(c) $5^{x+1} = 125$
(d) $\frac{27^{\frac{2}{3}} \times 64^{\frac{1}{2}}}{32^{\frac{2}{3}}}$
(e) $8^{\frac{1}{3}} \times 81^{\frac{1}{4}} \times 32^{\frac{1}{5}}$

2 Simplify the following:
(a) $3\sqrt{28} - 5\sqrt{63} + 4\sqrt{112}$
(b) $\frac{3\sqrt{8} \times 5\sqrt{5} \times \sqrt{7}}{\sqrt{42} \times 2\sqrt{3} \times \sqrt{15}}$
(c) $\frac{2}{2\sqrt{3} - \sqrt{7}}$
(d) $\frac{2\sqrt{3} - \sqrt{5}}{2\sqrt{3} + \sqrt{5}}$
(e) $log_5 0.04$

Answer :

Let's solve and simplify each given expression and equation step-by-step:

(c) $5^{x+1} = 125$

To solve for [tex]x[/tex]:


  1. Recognize that $125[tex]can be expressed as a power of $5[/tex]: $125 = 5^3$.

  2. The equation becomes $5^{x+1} = 5^3$.

  3. Since the bases are the same, set the exponents equal: [tex]x + 1 = 3[/tex].

  4. Solve for [tex]x[/tex]:
    [tex]x = 3 - 1 = 2[/tex]


(d) [tex]\frac{27^{\frac{2}{3}} \times 64^{\frac{1}{2}}}{32^{\frac{2}{3}}}[/tex]


  1. Simplify each part:

    • $27 = 3^3[tex], so $27^{\frac{2}{3}} = (3^3)^{\frac{2}{3}} = 3^{2} = 9[/tex].

    • $64 = 2^6[tex], so $64^{\frac{1}{2}} = (2^6)^{\frac{1}{2}} = 2^3 = 8[/tex].

    • $32 = 2^5[tex], so $32^{\frac{2}{3}} = (2^5)^{\frac{2}{3}} = 2^{\frac{10}{3}}[/tex].



  2. Substitute back:
    [tex]\frac{9 \times 8}{2^{\frac{10}{3}}}[/tex] = [tex]\frac{72}{2^{\frac{10}{3}}}[/tex]

  3. Simplify:

    • $72 = 2^3 \times 9$.

    • So, cancel out factors of 2 to further simplify the expression.




(e) $8^{\frac{1}{3}} \times 81^{\frac{1}{4}} \times 32^{\frac{1}{5}}$


  1. Simplify each part:

    • $8 = 2^3[tex], so $8^{\frac{1}{3}} = 2^{(3 \times \frac{1}{3})} = 2^{1} = 2[/tex].

    • $81 = 3^4[tex], so $81^{\frac{1}{4}} = 3^{(4 \times \frac{1}{4})} = 3^{1} = 3[/tex].

    • $32 = 2^5[tex], so $32^{\frac{1}{5}} = 2^{(5 \times \frac{1}{5})} = 2^{1} = 2[/tex].



  2. Combine:
    [tex]2 \times 3 \times 2 = 12[/tex]


(a) $3\sqrt{28} - 5\sqrt{63} + 4\sqrt{112}$


  1. Simplify each square root:

    • [tex]\sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7}[/tex].

    • [tex]\sqrt{63} = \sqrt{9 \times 7} = 3\sqrt{7}[/tex].

    • [tex]\sqrt{112} = \sqrt{16 \times 7} = 4\sqrt{7}[/tex].



  2. Substitute back:
    [tex]3(2\sqrt{7}) - 5(3\sqrt{7}) + 4(4\sqrt{7})[/tex]

  3. Combine like terms:
    [tex]6\sqrt{7} - 15\sqrt{7} + 16\sqrt{7} = 7\sqrt{7}[/tex]


(b) [tex]\frac{3\sqrt{8} \times 5\sqrt{5} \times \sqrt{7}}{\sqrt{42} \times 2\sqrt{3} \times \sqrt{15}}[/tex]


  1. Simplify:

    • [tex]\sqrt{8} = 2\sqrt{2}[/tex], [tex]\sqrt{42} = \sqrt{6 \times 7}[/tex], [tex]\sqrt{15} = \sqrt{3 \times 5}[/tex].



  2. Substitute & Reduce:

    • Numerator: $3 \times 2\sqrt{2} \times 5\sqrt{5} \times \sqrt{7}[tex]= $30\sqrt{70}[/tex]

    • Denominator simplifies to $6\sqrt{35}$.



  3. Simplify:
    [tex]\frac{30\sqrt{70}}{6\sqrt{35}} = 5\sqrt{2}[/tex]


(c) [tex]\frac{2}{2\sqrt{3} - \sqrt{7}}[/tex]


  1. Multiply by the conjugate:

    • [tex](2\sqrt{3} + \sqrt{7})/(2\sqrt{3} + \sqrt{7})[/tex]



  2. Depth:

    • Numerator: $2(2\sqrt{3} + \sqrt{7}) = 4\sqrt{3} + 2\sqrt{7}$

    • Denominator: [tex](2\sqrt{3})^2 - (\sqrt{7})^2 = 12 - 7 = 5[/tex]



  3. Result: [tex]\frac{4\sqrt{3} + 2\sqrt{7}}{5}[/tex]


(d) [tex]\frac{2\sqrt{3} - \sqrt{5}}{2\sqrt{3} + \sqrt{5}}[/tex]


  1. Multiply by the conjugate:

    • Numerator becomes $4 \times 3 - 5 = 12 - 5 = 7$



  2. Simplify:
    [tex]\frac{(2\sqrt{3} - \sqrt{5})(2\sqrt{3} - \sqrt{5})}{7} = -1[/tex]


(e) [tex]\log_5 0.04[/tex]


  1. Express $0.04[tex]as a fraction of the form $5^x[/tex]
    $0.04 = \frac{4}{100} = \frac{5^{-2}}{5^2}$

  2. Simplify to get the result for [tex]\log_5 0.04[/tex]
    $$x = -2\log_5{5} = -2"


Each step should help understand both the resolution and the simplification processes of the given problems.