Question
Solve $3 \times 9^x + 5 \times 3^x – 2 = 0$.
▶️Answer/Explanation
Solution: –
$3 \times (3^x)^2 + 5 \times 3^x – 2 = 0$
Let t = $3^x$ , therefore
$3 \times (t)^2 + 5 \times t – 2 = 0$
$3 \times (t)^2 + 6 \times t- 1 \times t – 2 = 0$
(3t – 1)(t+ 2) = 0, or
$(3 \times 3^x – 1)(3^x + 2) = 0 $
$3^x = \frac{1}{3} \quad (\text{or} \; 3^x = -2)$
$x = -1$
Question
It is given that \(log_{10}a=\frac{1}{3}\), where \(a> 0\).
Find the value of
(a) \(log_{10}(\frac{1}{a})\);
(b) \(log_{1000}a\)
▶️Answer/Explanation
Solution
Given that
\(log_{10}a\) = \(\frac{1}{3}\)
Applying the logarithm property
\(log_{10}\frac{X}{Y}\) = \(log_{10}X\) – \(log_{10}Y\)
put X= 1, Y= a, we get
(a) \(log_{10}\frac{1}{a}\) = \(log_{10}1-log_{10}a\)
Now, applying the property
\(log_{10}m^n{}\) = \(n log_{10}m{}\)
put m=a, n= -1, we get
\(log_{10}a^{-1}=-log_{10}a\)
putting the value of \(log_{10}a\) = \(\frac{1}{3}\)
we get \(log_{10}\frac{1}{a}\) =\(-\frac{1}{3}\)
(b) Applying the property \(log_{n}m\) = \(\frac{(log_{}m)}{(log_{}n)}\)
putting m= a, n= 1000 = 10^{3} and applying base 10, we get
\(\frac{log_{10}a}{log_{10}10^{3}}\) = \(\frac{log_{10}a}{3 log_{10}10^{}}\)
we know that \({log_{10}10^{}}\) = 1 and putting the value of \(log_{10}a\) = \(\frac{1}{3}\) we get
\(\frac{log_{10}a}{log_{10}10^{3}}\) = \(\frac{1}{3}\)\(\frac{(\frac{1}{3})}{1} \)
= \( \frac{1}{9}\)
Question
Solve the equation \(2 – {\log _3}(x + 7) = {\log _{\tfrac{1}{3}}}2x\) .
▶️Answer/Explanation
Markscheme
\({\log _3}\left( {\frac{9}{{x + 7}}} \right) = {\log _3}\frac{1}{{2x}}\) M1M1A1
Note: Award M1 for changing to single base, M1 for incorporating the 2 into a log and A1 for a correct equation with maximum one log expression each side.
\(x + 7 = 18x\) M1
\(x = \frac{7}{{17}}\) A1
[5 marks]
Question
The first terms of an arithmetic sequence are \(\frac{1}{{{{\log }_2}x}},{\text{ }}\frac{1}{{{{\log }_8}x}},{\text{ }}\frac{1}{{{{\log }_{32}}x}},{\text{ }}\frac{1}{{{{\log }_{128}}x}},{\text{ }} \ldots \)
Find x if the sum of the first 20 terms of the sequence is equal to 100.
▶️Answer/Explanation
Markscheme
METHOD 1
\(d = \frac{1}{{{{\log }_8}x}} – \frac{1}{{{{\log }_2}x}}\) (M1)
\( = \frac{{{{\log }_2}8}}{{{{\log }_2}x}} – \frac{1}{{{{\log }_2}x}}\) (M1)
Note: Award this M1 for a correct change of base anywhere in the question.
\( = \frac{2}{{{{\log }_2}x}}\) (A1)
\(\frac{{20}}{2}\left( {2 \times \frac{1}{{{{\log }_2}x}} + 19 \times \frac{2}{{{{\log }_2}x}}} \right)\) M1
\( = \frac{{400}}{{{{\log }_2}x}}\) (A1)
\(100 = \frac{{400}}{{{{\log }_2}x}}\)
\({\log _2}x = 4 \Rightarrow x = {2^4} = 16\) A1
METHOD 2
\({20^{{\text{th}}}}{\text{ term}} = \frac{1}{{{{\log }_{{2^{39}}}}x}}\) A1
\(100 = \frac{{20}}{2}\left( {\frac{1}{{{{\log }_2}x}} + \frac{1}{{{{\log }_{{2^{39}}}}x}}} \right)\) M1
\(100 = \frac{{20}}{2}\left( {\frac{1}{{{{\log }_2}x}} + \frac{{{{\log }_2}{2^{39}}}}{{{{\log }_2}x}}} \right)\) M1(A1)
Note: Award this M1 for a correct change of base anywhere in the question.
\(100 = \frac{{400}}{{{{\log }_2}x}}\) (A1)
\({\log _2}x = 4 \Rightarrow x = {2^4} = 16\) A1
METHOD 3
\(\frac{1}{{{{\log }_2}x}} + \frac{1}{{{{\log }_8}x}} + \frac{1}{{{{\log }_{32}}x}} + \frac{1}{{{{\log }_{128}}x}} + \ldots \)
\(\frac{1}{{{{\log }_2}x}} + \frac{{{{\log }_2}8}}{{{{\log }_2}x}} + \frac{{{{\log }_2}32}}{{{{\log }_2}x}} + \frac{{{{\log }_2}128}}{{{{\log }_2}x}} + \ldots \) (M1)(A1)
Note: Award this M1 for a correct change of base anywhere in the question.
\( = \frac{1}{{{{\log }_2}x}}(1 + 3 + 5 + \ldots )\) A1
\( = \frac{1}{{{{\log }_2}x}}\left( {\frac{{20}}{2}(2 + 38)} \right)\) (M1)(A1)
\(100 = \frac{{400}}{{{{\log }_2}x}}\)
\({\log _2}x = 4 \Rightarrow x = {2^4} = 16\) A1
[6 marks]