Show that \(a = 2\).
About this question
Question 5.2 of Mathematics Grade 12 Paper 1, NSC Examination (May/June) (DBE, May/June 2024). It tests the Term 1 ATP topic Exponential and Logarithmic Functions.
- TermTerm 1
- ATP topicExponential and Logarithmic Functions
- PaperPaper 1
- Marks2 marks for the method (formula and substitution) and the correct final answer
The graphs of \(f(x) = \log_a x\) and \(g\) are drawn. Graph \(g\) is the reflection of \(f\) in the line \(y = x\). Graph \(f\) passes through the point \(P(4\,;\,2)\). Q is the \(x\)-intercept of \(f\) and R is the \(y\)-intercept of \(g\).
Diagram: \(f\) is an increasing log curve that cuts the \(x\)-axis at Q (to the right of the origin O) and passes through \(P(4\,;\,2)\). \(g\) is the increasing exponential curve, which cuts the \(y\)-axis at R above O and rises steeply in the first quadrant.
Substitute \(P(4\,;\,2)\) into \(f(x) = \log_a x\):
\[2 = \log_a 4\]\[a^2 = 4\]\(a = 2\) (the base must be positive)
More Questions On Exponential and Logarithmic Functions
For the inverse swop \(x\) and \(y\):
\[x = \left(\frac{3}{2}\right)^y\]\(y = \log_{\frac{3}{2}} x\)
Swap \(x\) and \(y\):
\[x = \left(\frac{1}{3}\right)^y\]\(y = \log_{\frac{1}{3}} x\)
So \((4\,;\,3)\) lies on \(h\).
\(h\) is decreasing and always above its asymptote \(y = 0\).
\(0 \lt y \lt 3\)
The asymptote is \(y = -16\), so \(q = -16\).
\[f(x) = 2^{x + p} - 16\]Substitute \(T(3\,;\,16)\):
\[16 = 2^{3 + p} - 16\]\[2^{3 + p} = 32 = 2^5\]\[3 + p = 5\]\(p = 2\) and \(q = -16\)
\(f(x) = -1\):
\[\log_{4} x = -1 \quad\therefore x = 4^{-1} = \frac{1}{4}\]\(f(x) = 2\): \(x = 16\) (from 6.1)
\(f\) is increasing, so
\(\frac{1}{4} \lt x \lt 16\)