Question difficulty level:
Basic
Moderate
Challenging
Difficult
Genius :)
Basic
Moderate
Challenging
Difficult
Genius :)
Paper 1 – No Calculator
[Maximum mark: 7]
Solve \log_{2}(2x - 3) + \log_{2}(x + 5) = \log_{2}(3x - 3) + 2.
Using logarithmic laws we get:
\log_{2}(2x - 3) + \log_{2}(x + 5) = \log_{2}(3x - 3) + 2
\log_{2}((2x - 3)(x + 5)) = \log_{2}(3x - 3) + 2\log_{2}2
\log_{2}(2x^{2} + 7x - 15) = \log_{2}(3x - 3) + \log_{2}4
\log_{2}(2x^{2} + 7x - 15) = \log_{2}(4(3x - 3))
2x^{2} + 7x - 15 = 4(3x - 3)
Expanding, rearranging for zero and factorizing we get:
2x^{2} - 5x - 3 = 0
(2x + 1)(x - 3) = 0, therefore x = -\frac{1}{2} or x = 3
Checking the original equation we can see that if x = -\frac{1}{2}, then 2x - 3 < 0 and 3x - 3 < 0, hence:
\boldsymbol{{\color{Purple} x = 3}}
.
Paper 1 – No Calculator
[Maximum mark: 7]
Find the value of
(a) 27^{\log_{3}5}.
[3]
(b) \log_{6}144 + \log_{6}6 - 2\log_{6}2.
[4]
(a)
Using exponential and logarithmic laws we get:
27^{\log_{3}5} = \left ( 3^{3} \right )^{\log_{3}5} =
= 3^{3\log_{3}5} = 3^{\log_{3}5^{3}} = 3^{\log_{3}125} =
\boldsymbol{{\color{Purple} = 125}}
(b)
Using logarithmic laws and rearranging the terms we get:
\log_{6}144 + \log_{6}6 - 2\log_{6}2 = \log_{6}144 - \log_{6}2^{2} + \log_{6}6 =
= \log_{6}144 - \log_{6}4 + 1 = \log_{6}\left ( \frac{144}{4} \right ) + 1 =
=\log_{6}36 + 1 = \log_{6}6^{2} + 1 = 2\log_{6}6 + 1 = 2 + 1 =
\boldsymbol{{\color{Purple} 3}}
.
Paper 1 – No Calculator
[Maximum mark: 5]
Solve the equation \ln8 = 3\ln x - 12.
Give your answer in the form pe^{q} where p and q are positive integers.
Rearranging and using logarithmic laws, we get:
\ln x^{3} - \ln8 = 12
\ln \left ( \frac{x^{3}}{8} \right ) = 12
Changing the equation from logarithmic to exponential form, we get:
\frac{x^{3}}{8} = e^{12}
Solving for x gives us that:
x = \sqrt[3]{8e^{12}}, hence:
\boldsymbol{{\color{Purple} x = 2e^{4}}}
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Paper 1 – No Calculator
[Maximum mark: 6]
Solve 4^{x} = 3 - 2^{x + 2}. Give your answer in the form \log_{p}(q + \sqrt{r}), where p,q,r\, \epsilon \, \mathbb{Z}.
Using exponent laws and rearranging we get that:
\left ( 2^{2} \right )^{x} = 3 - 2^{2} \times 2^{x}
\left ( 2^{x} \right )^{2} + 4 \times 2^{x} - 3 = 0
Using the substitution a = 2^{x}, we get:
a^{2} + 4a - 3 = 0
Using the quadratic formula to solve for a:
a = \frac{-4 \pm \sqrt{4^{2} - 4(1)(-3)}}{2(1)} = \frac{-4 \pm \sqrt{28}}{2} = \frac{-4 \pm 2\sqrt{7}}{2} = -2 \pm \sqrt{7}
Since 2^{x} > 0, we get that:
2^{x} = -2 + \sqrt{7}
Changing from an exponential to a logarithmic form:
\boldsymbol{{\color{Purple} x = \log_{2}(-2 + \sqrt{7})}}
.
Paper 1 – No Calculator
[Maximum mark: 8]
Consider function g(x) = \log_{(p-2)}(x^{2} - 4x + 13) where p>2 and p \neq 3. There is exactly one point on the graph of g where x = 2.
Find the value of p.
The information that there is exactly one point on the graph of g where x = 2 means that the equation g(x) = 2 has exactly one solution.
So we will start by writing that:
\log_{(p-2)}(x^{2} - 4x + 13) = 2
Converting from logarithmic to exponential form and working further we get that:
(p - 2)^{2} = x^{2} - 4x + 13
p^{2} - 4p + 4 = x^{2} - 4x + 13
x^{2} - 4x - p^{2} + 4p + 9 = 0
This equation still has exactly one solution, therefore its discriminant is equal to zero.
Hence we get that:
(-4)^{2} - (4)(1)(- p^{2} + 4p + 9) = 0
p^{2} - 4p - 5 = 0
(p + 1)(p - 5) = 0
p = -1 or p = 5
Since p>2, the final answer is:
\boldsymbol{{\color{Purple} p = 5}}
.
Paper 1 – No Calculator
[Maximum mark: 5]
In this question, give all your answers as an integer.
Find the value of
(a) \log_{4}4.
[1]
(b) \log_{4}2 + \log_{4}32.
[2]
(c) \log_{4}48 - \log_{4}3.
[2]
(a)
The base and the number inside the logarithm are the same, so:
\boldsymbol{{\color{Purple} \log_{4}4 = 1}}
(b)
Using logarithmic laws we get that:
\log_{4}2 + \log_{4}32 = \log_{4}(2 \times 32) = \log_{4}64
Rewriting the expression and using logarithmic laws, we get:
\log_{4}64 = \log_{4}4^{3} = 3\log_{4}4, therefore:
\boldsymbol{{\color{Purple} \log_{4}2 + \log_{4}32 = 3}}
(c)
Using logarithmic laws we get that:
\log_{4}48 - \log_{4}3 = \log_{4}(\frac{48}{3}) = \log_{4}16
Rewriting the expression and using logarithmic laws, we get:
\log_{4}16 = \log_{4}4^{2} = 2\log_{4}4, therefore:
\boldsymbol{{\color{Purple} \log_{4}48 - \log_{4}3 = 2}}
.