Solving Simultaneous Logarithmic Equations

When simultaneous equations are linear in the logarithms, treat $\log_2 x$ and $\log_2 y$ as the unknowns, solve as an ordinary linear pair, then convert back with the definition of a log.

Solve the simultaneous equations $\log_2 x + \log_2 y = 5$ $\log_2 x - \log_2 y = 1$ [5]

The working

Step 1, treat $\log_2 x$ and $\log_2 y$ as variables and add the equations: $2\log_2 x = 6 \implies \log_2 x = 3 \quad \text{(M1, A1)}$

Step 2, subtract to find the other: $2\log_2 y = 4 \implies \log_2 y = 2 \quad \text{(A1)}$

Step 3, convert each log back to a value using $\log_2 x = 3 \Leftrightarrow x = 2^3$: $x = 2^3 = 8, \qquad y = 2^2 = 4 \quad \text{(A1, A1)}$

Check: $\log_2 8 + \log_2 4 = 3 + 2 = 5$ and $3 - 2 = 1$. ✓

Where the marks are won and lost

• Solve for $\log_2 x$ and $\log_2 y$ first, then convert. Rushing to $x$ and $y$ before the logs are found tangles the algebra.
• Convert correctly: $\log_2 x = 3$ means $x = 2^3 = 8$, not $x = 3 \times 2$.
• A quick check in the original equations catches conversion slips cheaply.

Common mistakes

• Writing $\log_2 x = 3 \implies x = 6$ (multiplying instead of raising the base to the power).
• Combining $\log_2 x + \log_2 y$ into $\log_2(xy)$ prematurely, which complicates the elimination.
• Losing a factor of $2$ when adding or subtracting.

Topic home: Logarithmic & Exponential Functions pillar. More: Worked examples.