Year 11 (2026) Maths Advanced (2026) Logs and Exponentials

Natural Logarithms

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Questions

Question 1

124514

Solve \(\log _e(2 x-1)-\log _e x=1\)

Answer

\(\dfrac{1}{2-e}\)

Worked Solution

\begin{aligned}
\log _e(2 x-1)-\log _e x & =1 \\
\log _e \frac{2 x-1}{x} & =1 \\
\frac{2 x-1}{x} & =e \\
2 x-1 & =e x \\
2 x-e x & =1 \\
x(2-e) & =1 \\
\therefore x & =\frac{1}{2-e}
\end{aligned}

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