Elite Mathematics / Linear / Topics / Algebraic Roots & Indices IGCSE · 4MA1 · Topic Algebraic Roots & Indices: past paper questions & worked solutions
Every Algebraic Roots & Indices question from the 4MA1 papers in our bank, 2020–2026, sorted from easiest to hardest. Try each one first, then open the worked solution.
49 questions
191 marks in total
70% of papers (38 of 54) include it
11/26/12 easy / medium / hard
Questions, easiest first Level is estimated from marks: 1–2 easy, 3–4 medium, 5+ hard.
1. Jun 2026 P1HR · Q3 Easy 2 marksShow worked solution Part (a) | Step 1 \[e^7\times e^3=e^{7+3}=e^{10}.\]
For equal bases, add the indices when multiplying.
Part (b) | Step 1 \[g^5\div g^2=g^{5-2}=g^3\quad(g\ne0).\]
For equal bases, subtract the indices when dividing.
Final answer (a) \(e^{10}\)
(b) \(g^3\)
2. Jun 2026 P1H · Q6 Easy 2 marksShow worked solution Solution | Step 1 \[(4x^5y^6)^3=4^3x^{5\times3}y^{6\times3}=64x^{15}y^{18}.\]
Raise each factor to the third power and multiply exponents for a power of a power.
Final answer \(64x^{15}y^{18}\)
3. Jun 2026 P2HR · Q16 Easy 2 marksShow worked solution Solution | Step 1 \[\frac1{\sqrt a}=a^{-1/2},\qquad\left(a^{-1/2}\right)^6=a^{-3}.\]
A reciprocal changes the sign of the exponent and an outer power multiplies exponents.
4. May 2026 4WM1H · Q6 Easy 2 marksShow worked solution Cube every part \[(4x^5y^6)^3=4^3x^{15}y^{18}=64x^{15}y^{18}\]
Final answer \(64x^{15}y^{18}\)
5. Nov 2025 4WM1H · Q7 Easy 2 marksShow worked solution Use index laws \[
\frac{8^{-2}\times8^9}{8^{10}}=8^{-2+9-10}=8^{-3}
\]
6. Nov 2025 P1H · Q8 Easy 2 marksShow worked solution Use index laws \[
\frac{8^{-2}\times8^9}{8^{10}}=8^{-2+9-10}=8^{-3}
\]
7. Jun 2024 P2HR · Q7 Easy 2 marksShow worked solution Use index laws \[
\frac{3^{-2}\times 3^5}{3^{10}}=3^n
\]
Use index laws Use index laws:
\[
\frac{3^{-2+5}}{3^{10}}=3^{3-10}=3^{-7}
\]
Use index laws So
\[
n=-7
\]
8. Jun 2023 P2HR · Q7 Easy 2 marksShow worked solution Use index laws Any non-zero expression to the power 0 is 1:
\[
(4^{-2})^0=1
\]
Use index laws For part (b),
\[
3^{-14}\times 3^8=3^{-14+8}=3^{-6}
\]
Use index laws So
\[
m=-6
\]
Final answer (a) \(1\), (b) \(m=-6\).
9. Jun 2023 P2HR · Q11 Easy 2 marksShow worked solution Convert standard form \[
\left(\frac{m^6k^{10}}{25}\right)^{\frac32}
=\frac{(m^6)^{\frac32}(k^{10})^{\frac32}}{25^{\frac32}}
\]
\[
(m^6)^{\frac32}=m^9
\]
\[
(k^{10})^{\frac32}=k^{15}
\]
Convert standard form Also,
\[
25^{\frac32}=(\sqrt{25})^3=5^3=125
\]
Convert standard form So
\[
\left(\frac{m^6k^{10}}{25}\right)^{\frac32}
=\frac{m^9k^{15}}{125}
\]
Final answer \(\displaystyle \frac{m^9k^{15}}{125}\)
10. Jun 2022 P2HR · Q7 Easy 2 marksShow worked solution Use index laws \[
\frac{2^k}{4^n}=2^x
\]
Use index laws Since \(4=2^2\),
\[
4^n=(2^2)^n=2^{2n}
\]
Use index laws So
\[
\frac{2^k}{2^{2n}}=2^{k-2n}
\]
Use index laws Therefore
\[
x=k-2n
\]
11. Specimen 4WM2H · Q3 Easy 2 marksShow worked solution Use index laws \[
4^n=(2^2)^n=2^{2n}
\]
Use index laws So
\[
\frac{2^k}{4^n}=\frac{2^k}{2^{2n}}=2^{k-2n}
\]
Use index laws Since this equals \(2^x\),
\[
x=k-2n
\]
12. Jun 2026 P1HR · Q26 Medium 3 marksShow worked solution Solution | Step 1 \[27=3^3,\qquad9=3^2,\qquad\frac{6^x}{2^x}=\left(\frac62\right)^x=3^x.\]
Write both sides in a common base.
Solution | Step 2 \[\frac{27^x\times3^{n+6}}{9^{4-n}}=3^{3x+n+6-2(4-n)}=3^{3x+3n-2}.\]
Multiply powers by adding indices and divide powers by subtracting indices.
Solution | Step 3 \[3^{3x+3n-2}=3^x\quad\Longrightarrow\quad3x+3n-2=x.\]
The base-3 exponential is one-to-one, so equal powers have equal indices.
Solution | Step 4 \[2x=2-3n\quad\Longrightarrow\quad x=1-\frac{3n}{2}.\]
Collect the x terms and divide by 2.
Final answer \(x=1-\frac{3n}{2}\)
13. Nov 2025 4WM1H · Q15 Medium 3 marksShow worked solution Evaluate fraction \[
\frac{16^{x-3}\times 8^{x+1}}{32^x}
=\frac{2^{4x-12}\times 2^{3x+3}}{2^{5x}}
\]
\[
=2^{4x-12+3x+3-5x}=2^{2x-9}
\]
14. Jun 2025 P2H · Q15 Medium 3 marksShow worked solution Use index laws \[
\left(\frac{2a^3}{10a^7c^2}\right)^{-3}
\]
Use index laws First simplify inside the bracket:
\[
\frac{2a^3}{10a^7c^2}=\frac{1}{5a^4c^2}
\]
A negative power means take A negative power means take the reciprocal:
\[
\left(\frac{1}{5a^4c^2}\right)^{-3}=(5a^4c^2)^3
\]
\[
=125a^{12}c^6
\]
Final answer \(125a^{12}c^6\).
15. May 2025 4WM1H · Q9 Medium 3 marksShow worked solution Use index laws \[
7^x=1
\]
Use index laws Since \(7^0=1\),
\[
x=0
\]
Use index laws For part (b),
\[
\frac{8^{-4}\times8^{11}}{8^{12}}=8^n
\]
\[
8^{-4}\times8^{11}=8^7
\]
\[
\frac{8^7}{8^{12}}=8^{7-12}=8^{-5}
\]
Use index laws So \(n=-5\).
Final answer \(x=0\), \(n=-5\).
16. May 2025 4WM1H · Q16 Medium 3 marksShow worked solution Solve equation \[
8^{3y+4}\times4^{3y}=2^{5y}
\]
Solve equation Write \(8\) and \(4\) as powers of \(2\).
\[
(2^3)^{3y+4}(2^2)^{3y}=2^{5y}
\]
\[
2^{9y+12}\times2^{6y}=2^{5y}
\]
\[
2^{15y+12}=2^{5y}
\]
Solve equation So
\[
15y+12=5y
\]
\[
y=-\frac65
\]
Final answer \(y=-\frac65\)
17. Nov 2024 P2H · Q14 Medium 3 marksShow worked solution Use index laws \[
\frac{3^{2n+3}}{3^4}=3^{2n-1}
\]
\[
3^3\times3^{1-2n}=3^{4-2n}
\]
Use index laws So \[ 2n-1=4-2n \]
\[
4n=5
\]
Final answer \(n=\frac54\)
18. Jun 2024 P2H · Q23 Medium 3 marksShow worked solution Simplify surd \[
\frac{30\times 25^{2x+7}}{\sqrt{180}\times (\sqrt5)^{4x+9}}
\]
Simplify the number part Simplify the number part:
\[
\sqrt{180}=6\sqrt5
\]
\[
\frac{30}{6\sqrt5}=\frac{5}{\sqrt5}=\sqrt5=5^{1/2}
\]
Simplify surd Now write the powers with base 5:
\[
25^{2x+7}=(5^2)^{2x+7}=5^{4x+14}
\]
\[
(\sqrt5)^{4x+9}=5^{(4x+9)/2}=5^{2x+9/2}
\]
Use index laws So the full expression is
\[
5^{1/2+4x+14-(2x+9/2)}
\]
\[
=5^{2x+10}
\]
Final answer \(5^{2x+10}\).
19. Jun 2022 P2H · Q24 Medium 3 marksShow worked solution Simplify surd \[
\frac{18(\sqrt{27})^{4n+6}}{6\times 9^{2n+8}}=3^x
\]
Use index laws First,
\[
\frac{18}{6}=3=3^1
\]
Simplify surd Also,
\[
\sqrt{27}=\sqrt{3^3}=3^{3/2}
\]
Simplify surd So
\[
(\sqrt{27})^{4n+6}=3^{\frac32(4n+6)}=3^{6n+9}
\]
Use index laws And
\[
9^{2n+8}=(3^2)^{2n+8}=3^{4n+16}
\]
Use index laws Therefore the whole expression is
\[
3^{1+6n+9-(4n+16)}=3^{2n-6}
\]
Use index laws So
\[
x=2n-6
\]
20. May 2021 P2H · Q14 Medium 3 marksShow worked solution Use index laws \[
\left(\frac{9x^4}{16y^{10}}\right)^{-1/2}
\]
A negative power takes the A negative power takes the reciprocal:
\[
\left(\frac{16y^{10}}{9x^4}\right)^{1/2}
\]
Simplify surd Now take the square root of each factor:
\[
\frac{\sqrt{16y^{10}}}{\sqrt{9x^4}}=\frac{4y^5}{3x^2}
\]
Final answer \(\frac{4y^5}{3x^2}\).
21. Jan 2021 P2H · Q11 Medium 3 marksShow worked solution Use index laws \[
\left(\frac{9t^4w^9}{18t^6w^{10}}\right)^{-2}
\]
Simplify inside the brackets Simplify inside the brackets:
\[
\frac{9t^4w^9}{18t^6w^{10}}
=\frac{1}{2}t^{-2}w^{-1}
=\frac{1}{2t^2w}
\]
Use index laws Now apply the power \(-2\):
\[
\left(\frac{1}{2t^2w}\right)^{-2}
=(2t^2w)^2
\]
\[
=4t^4w^2
\]
22. Jan 2020 P2H · Q1 Medium 3 marksShow worked solution (a) \[
\frac{x^9}{x^2}=x^{9-2}=x^7
\]
(b) \[
\frac{7^8\times 7^4}{7^3}=7^{8+4-3}=7^9
\]
Final answer (a) \(x^7\), (b) \(7^9\).
23. Specimen 4WM1H · Q22 Medium 3 marksShow worked solution Write everything as powers of 3 \[
\sqrt{27}=3^{3/2},\qquad 9=3^2,\qquad \frac{18}{6}=3
\]
Simplify the powers \[
\frac{18(\sqrt{27})^{4n+6}}{6(9)^{2n+8}}
=3\times \frac{\left(3^{3/2}\right)^{4n+6}}{(3^2)^{2n+8}}
\]
\[
=3^1\times \frac{3^{6n+9}}{3^{4n+16}}
\]
Subtract the indices \[
3^{1+6n+9-(4n+16)}=3^{2n-6}
\]
So
\[
x=2n-6
\]
24. Nov 2025 P2H · Q13 Medium 4 marksShow worked solution (a) \[
\left(\frac{125x^6}{y^{15}}\right)^{1/3}
=\frac{5x^2}{y^5}
\]
(b) \[
64^p=(4^3)^p=4^{3p}
\]
\[
4^n=\frac{4^m}{64^p}=\frac{4^m}{4^{3p}}=4^{m-3p}
\]
Final answer (a) \(\frac{5x^2}{y^5}\), (b) \(n=m-3p\)
25. Jun 2025 P1H · Q4 Medium 4 marksShow worked solution Use index laws For part (a),
\[
x^5\times x^7=x^{12}
\]
Use index laws So \(m=12\).
Use index laws For part (b),
\[
y^8\div y^3=y^{8-3}=y^5
\]
Use index laws So \(n=5\).
Use index laws For part (c),
\[
(5a^4r^2)^3=5^3a^{4\times 3}r^{2\times 3}
\]
\[
=125a^{12}r^6
\]
Final answer (a) \(m=12\), (b) \(n=5\), (c) \(125a^{12}r^6\).
26. May 2025 4WM1H · Q5 Medium 4 marksShow worked solution Use index laws \[
x^5\times x^7=x^{5+7}=x^{12}
\]
Use index laws So \(m=12\).
\[
y^8\div y^3=y^{8-3}=y^5
\]
Use index laws So \(n=5\).
\[
(5a^4r^2)^3=5^3(a^4)^3(r^2)^3=125a^{12}r^6
\]
Final answer \(m=12\), \(n=5\), and \(125a^{12}r^6\).
27. May 2025 4WM1HR · Q5 Medium 4 marksShow worked solution Evaluate fraction \[
3^8\times 3^4 = 3^{8+4}=3^{12}
\]
\[
\frac{4^{12}}{4^7}=4^{12-7}=4^5
\]
\[
(6^2)^{10}=6^{20}
\]
\[
5^0=1
\]
Final answer \(m=12\), \(n=5\), \(p=20\), \(r=0\).
28. May 2024 P1H · Q10 Medium 4 marksShow worked solution (a) (a) \((2p)^0=1\).
Use index laws (b) \(y^9\times y^{-3}=y^{9-3}=y^6\), so \(n=6\).
(c) (c) \((5a^4c^2)^3=5^3a^{12}c^6=125a^{12}c^6\).
Final answer (a) \(1\), (b) \(6\), (c) \(125a^{12}c^6\)
29. May 2024 P1HR · Q17 Medium 4 marksShow worked solution (a) so (a) \[ \left(\sqrt[4]{k^{12}}\right)^5=(k^3)^5=k^{15} \] so \(n=15\).
(b) \[
\frac{7}{2-\sqrt3}\times\frac{2+\sqrt3}{2+\sqrt3}
=\frac{7(2+\sqrt3)}{4-3}=14+7\sqrt3
\]
\[
7\sqrt3=\sqrt{49\times3}=\sqrt{147}
\]
Final answer (a) \(15\), (b) \(\sqrt{147}+14\)
30. Nov 2023 P2H · Q12 Medium 4 marksShow worked solution (a) \[
2^7\times4^5=2^7\times(2^2)^5=2^{17}
\]
Use index laws Also \[ 4^x=(2^2)^x=2^{2x} \]
Use index laws So \[ 2x=17 \]
\[
x=\frac{17}{2}
\]
(b) \[
(125p^6y^{24})^{2/3}=125^{2/3}p^4y^{16}
\]
\[
=25p^4y^{16}
\]
Final answer (a) \(x=\frac{17}{2}\), (b) \(25p^4y^{16}\)
31. May 2023 P1HR · Q6 Medium 4 marksShow worked solution (a) (a) \(m^{10}\div m^3=m^{10-3}=m^7\).
Use index laws (b) \(k^n\times k^4=k^{n+4}=k^{12}\), so \(n+4=12\) and \(n=8\).
(c) (c) \((3x^6y^8)^2=3^2x^{12}y^{16}=9x^{12}y^{16}\).
Final answer (a) \(m^7\), (b) \(8\), (c) \(9x^{12}y^{16}\)
32. Jun 2022 P1H · Q9 Medium 4 marksShow worked solution Use index laws \[
8\times (4t)^0=8\times 1=8
\]
Use index laws For part (b),
\[
x^6\div x^{-5}=x^{6-(-5)}=x^{11}
\]
Use index laws So
\[
p=11
\]
Use index laws For part (c),
\[
(2k^2m^4)^3=2^3k^{2\times 3}m^{4\times 3}
\]
\[
=8k^6m^{12}
\]
Final answer (a) \(8\), (b) \(p=11\), (c) \(8k^6m^{12}\).
33. Jan 2022 P2HR · Q6 Medium 4 marksShow worked solution Use index laws \[
x^0=1
\]
Use index laws For part (b),
\[
2^{-3}\times 2^9=2^{-3+9}=2^6
\]
Use index laws So
\[
n=6
\]
Use index laws For part (c),
\[
\frac{7^{206}\times 7^m}{7^{214}}=7^{206+m-214}
\]
Use index laws This equals \(7^{-3}\), so
\[
206+m-214=-3
\]
\[
m=5
\]
Final answer (a) \(1\), (b) \(n=6\), (c) \(m=5\).
34. Nov 2021 P2H · Q1 Medium 4 marksShow worked solution Use index laws (a) \(3^4\times3^5=3^9\), so \(m=9\).
Use index laws (b) \((5^3)^7=5^{21}\), so \(n=21\).
(c) \[
\frac{7^8\times7^2}{7^p}=7^{10-p}=7^6
\]
Use index laws So \(10-p=6\), hence \(p=4\).
Final answer \(m=9\), \(n=21\), \(p=4\)
35. May 2021 P1H · Q11 Medium 4 marksShow worked solution Simplify surd \[
\sqrt2\times 16=2^x
\]
Write both factors as powers Write both factors as powers of 2:
\[
\sqrt2=2^{1/2}, \quad 16=2^4
\]
Use index laws So
\[
2^{1/2}\times 2^4=2^{9/2}
\]
Use index laws Therefore
\[
x=\frac92
\]
Use index laws For part (b),
\[
\frac{(11^{-6})^5}{11^4}=\frac{11^{-30}}{11^4}=11^{-30-4}=11^{-34}
\]
Use index laws So
\[
n=-34
\]
Final answer (a) \(x=\frac92\), (b) \(n=-34\).
36. Jan 2021 P2HR · Q4 Medium 4 marksShow worked solution Use index laws For part (a),
\[
(3k^2)^4=3^4(k^2)^4
\]
\[
=81k^8
\]
Use index laws For part (b),
\[
(21m^4n)\div(3n^{-5})=\frac{21m^4n}{3n^{-5}}
\]
\[
=7m^4n^{1-(-5)}
\]
\[
=7m^4n^6
\]
Final answer (a) \(81k^8\), (b) \(7m^4n^6\).
37. Specimen 4WM1H · Q7 Medium 4 marksShow worked solution Use the zero power \[
(4t)^0=1
\]
\[
8(4t)^0=8
\]
Subtract the indices \[
x^6\div x^{-5}=x^{6-(-5)}=x^{11}
\]
So \(p=11\).
Raise each factor to the power \[
(2k^2m^4)^3=2^3k^{2\times 3}m^{4\times 3}
\]
\[
=8k^6m^{12}
\]
Final answer \(8\), \(p=11\), and \(8k^6m^{12}\).
38. Jun 2025 P2HR · Q6 Hard 5 marksShow worked solution Use index laws \[
5^0=1
\]
Use index laws For part (b),
\[
\frac{5^9\times 5^{-3}}{5^{-2}}=5^{9+(-3)-(-2)}=5^8
\]
Use index laws So
\[
k=8
\]
Use index laws For part (c),
\[
(2d^4e^5)^3=2^3d^{4\times 3}e^{5\times 3}
\]
\[
=8d^{12}e^{15}
\]
Final answer (a) \(1\), (b) \(k=8\), (c) \(8d^{12}e^{15}\).
39. May 2023 P1HR · Q24 Hard 5 marksShow worked solution Use index laws \[
16^x=2^{4x}, \qquad 8=2^3
\]
Use index laws So \[ 2^n=2^{x^2}\times2^{4x}\times2^3=2^{x^2+4x+3} \]
Hence \[
n=x^2+4x+3=(x+2)^2-1
\]
\[
(x+2)^2=n+1
\]
Simplify surd Since \(x>0\), \[ x+2=\sqrt{n+1} \]
\[
x=-2+\sqrt{n+1}
\]
Simplify surd For \(x>0\), \(\sqrt{n+1}>2\), so \(n>3\).
Final answer \(x=-2+\sqrt{n+1}\), \(n>3\)
40. May-Nov 2020 P1HR · Q21 Hard 5 marksShow worked solution Use index laws \[
18^{4n}=(2\times3^2)^{4n}=2^{4n}3^{8n}
\]
\[
12^2=(2^2\times3)^2=2^4 3^2
\]
Use index laws So \[ M=\frac{2^{4n}3^{8n}\times2^{3(n^2-6n)}\times3^{2(1-4n)}}{2^4 3^2} \]
Combine powers of 2 Combine powers of 2: \[ 2^{4n+3n^2-18n-4}=2^{3n^2-14n-4} \]
Combine powers of 3 Combine powers of 3: \[ 3^{8n+2-8n-2}=3^0=1 \]
Use index laws So \[ M=2^{3n^2-14n-4} \]
Use index laws Given \(M=2\), \[ 3n^2-14n-4=1 \]
\[
3n^2-14n-5=0
\]
\[
(3n+1)(n-5)=0
\]
Final answer \(n=-\frac13\) or \(n=5\)
41. May 2026 4WM1H · Q12 Hard 6 marksShow worked solution (a) Write everything as powers of 2 \[8^x=2^{3x},\quad 64=2^6\Rightarrow\frac{2^{x/3}}{2^{3x}}=2^6\]
Compare powers \[\frac{x}{3}-3x=6\Rightarrow -\frac{8x}{3}=6\Rightarrow x=-\frac94\]
(b) Expand two brackets first \[(t+5)(2t-3)=2t^2+7t-15\]
Multiply by 4t \[4t(2t^2+7t-15)=8t^3+28t^2-60t\]
Final answer (a) \(x=-\frac94\) (b) \(8t^3+28t^2-60t\)
42. Jan 2022 P1H · Q1 Hard 6 marksShow worked solution (a) \[
a^7\times a^4=a^{11}
\]
(b) \[
w^{15}\div w^3=w^{12}
\]
(c) \[
(8x^5y^3)^2=64x^{10}y^6
\]
(d) \[
c=t^3-8v
\]
\[
t^3=c+8v
\]
Final answer \(a^{11}\), \(w^{12}\), \(64x^{10}y^6\), \(t=\sqrt[3]{c+8v}\)
43. Jan 2020 P1H · Q17 Hard 6 marksShow worked solution Simplify surd First,
\[
\sqrt{12}=2\sqrt3
\]
Simplify surd So
\[
6+2\sqrt{12}=6+4\sqrt3
\]
Simplify surd Now square:
\[
(6+4\sqrt3)^2=36+48\sqrt3+48
\]
\[
=84+48\sqrt3
\]
Factor out Factor out \(12\):
\[
84+48\sqrt3=12(7+4\sqrt3)
\]
Simplify surd For part (b),
\[
\left(\frac{27a^{12}}{t^{15}}\right)^{-\frac23}
=\left(\frac{t^{15}}{27a^{12}}\right)^{\frac23}
\]
\[
=\frac{t^{10}}{9a^8}
\]
Final answer (a) shown, (b) \(\displaystyle \frac{t^{10}}{9a^8}\)
44. Specimen 4WM1H · Q13 Hard 6 marksShow worked solution Multiply the conjugate factors \[
(3x-1)(3x+1)=9x^2-1
\]
Expand the remaining bracket \[
(9x^2-1)(x+2)=9x^3+18x^2-x-2
\]
Simplify inside the bracket \[
\frac{2x^5}{8xy^2}=\frac{x^4}{4y^2}
\]
Apply the negative power \[
\left(\frac{x^4}{4y^2}\right)^{-2}
=\left(\frac{4y^2}{x^4}\right)^2
=\frac{16y^4}{x^8}
\]
Final answer \(9x^3+18x^2-x-2\), and \(\frac{16y^4}{x^8}\).
45. Jun 2023 P2H · Q15 Hard 7 marksShow worked solution Solve equation For part (a),
\[
(32a^{15})^{3/5}=32^{3/5}a^{15\times 3/5}
\]
\[
=8a^9
\]
Solve equation For part (b),
\[
\left(\frac{1}{10x}\right)^{-3}=(10x)^3=1000x^3
\]
Solve equation For part (c),
\[
\frac{1-2y}{3}=\frac45-\frac{2y-1}{2}
\]
Multiply by Multiply by \(30\):
\[
10(1-2y)=24-15(2y-1)
\]
\[
10-20y=39-30y
\]
\[
10y=29
\]
\[
y=\frac{29}{10}
\]
Final answer (a) \(8a^9\), (b) \(1000x^3\), (c) \(y=\frac{29}{10}\)
46. Jan 2021 P2HR · Q19 Hard 7 marksShow worked solution Simplify surd \[
8^2=(2^3)^2=2^6
\]
\[
\sqrt[3]{4^6}=\sqrt[3]{(2^2)^6}=\sqrt[3]{2^{12}}=2^4
\]
Simplify surd So
\[
8^2\times \sqrt[3]{4^6}=2^6\times 2^4=2^{10}
\]
Use index laws For part (b),
\[
n^{-4/5}=\left(\frac12\right)^4=2^{-4}
\]
Use index laws So
\[
n^{4/5}=2^4
\]
Raise both sides to the Raise both sides to the power \(\frac54\):
\[
n=(2^4)^{5/4}=2^5=32
\]
Final answer (a) \(2^{10}\), (b) \(n=32\).
47. Jan 2020 P2HR · Q21 Hard 7 marksShow worked solution Factorise expression For part (a), factorise the numerator and denominator:
\[
10x^2+23x+12=(5x+4)(2x+3)
\]
\[
4x^2-9=(2x-3)(2x+3)
\]
Use index laws So
\[
\frac{10x^2+23x+12}{4x^2-9}=\frac{(5x+4)(2x+3)}{(2x-3)(2x+3)}
\]
\[
=\frac{5x+4}{2x-3}
\]
Use index laws For part (b),
\[
2^{2y}\times 2^{3y+2}=2^{5y+2}
\]
Use index laws Also,
\[
\frac{8^{5y}}{4^n}=\frac{(2^3)^{5y}}{(2^2)^n}=2^{15y-2n}
\]
Equate powers of 2 Equate powers of 2:
\[
5y+2=15y-2n
\]
\[
2n=10y-2
\]
\[
n=5y-1
\]
Final answer (a) \(\frac{5x+4}{2x-3}\), (b) \(n=5y-1\).
48. Nov 2023 P2H · Q4 Hard 8 marksShow worked solution (a) \[
\frac{2x+5}{6}=2x-5
\]
\[
2x+5=12x-30
\]
\[
35=10x
\]
\[
x=\frac72
\]
(b) \[
h^{15}\div h^3=h^{15-3}=h^{12}
\]
(c) \[
(2g^3k^5)^4=2^4g^{12}k^{20}=16g^{12}k^{20}
\]
(d) \[
\frac{y^5\times y^n}{y^7}=y^{5+n-7}=y^{n-2}
\]
Use index laws Given that this equals \(y^{12}\), \[ n-2=12 \]
\[
n=14
\]
Final answer \(x=\frac72\), \(h^{12}\), \(16g^{12}k^{20}\), and \(n=14\).
49. Jan 2021 P1HR · Q12 Hard 9 marksShow worked solution (a) \[
(16e^{10}f^6)^{\frac12}
=4e^5f^3
\]
(b) \[
\frac{2x+1}{4}+\frac{x-2}{3}
=\frac{3(2x+1)+4(x-2)}{12}
\]
\[
=\frac{6x+3+4x-8}{12}
=\frac{10x-5}{12}
=\frac{5(2x-1)}{12}
\]
(c) \[
4^{k+3}=16\times 2^k
\]
\[
(2^2)^{k+3}=2^4\times 2^k
\]
\[
2^{2k+6}=2^{k+4}
\]
\[
2k+6=k+4
\]
\[
k=-2
\]
Final answer (a) \(4e^5f^3\), (b) \(\frac{5(2x-1)}{12}\), (c) \(k=-2\).
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