Mathematics 9709/32 — May/June 2010

Walkthrough

We need to solve for $x$ when $x$ appears in an exponent inside a fraction. The key is to treat $2^x$ as a single unknown. Multiply both sides by $2^x - 1$ to clear the denominator. This gives $2^x + 1 = ​​5(2^x - ​​1)$. Expanding the right-hand side gives $2^x + 1 = ​​5 \cdot ​​2^x - ​​5$. Rearranging to get all $2^x$ terms together: $1 + ​​5 = ​​5 \cdot ​​2^x - ​​2^x$, so $6 = ​​4 \cdot ​​2^x$. Dividing by 4 gives $2^x = ​​\frac{3}{2}$. Now the variable is in the exponent. Taking log base 2 of both sides gives $x = ​​\log_2\left(\frac{3}{2}\right)$. Use the change-of-base formula to evaluate with a calculator: $x = ​​\frac{\ln(3/2)}{\ln(2)} = ​​0.584962\ldots$. Rounding to 3 significant figures gives $0.585$.

Key Takeaways

• An equation with $2^x$ can be solved by isolating $2^x$ first, then applying logarithms.
• The change-of-base formula $\log_2(a) = ​​\frac{\ln(a)}{\ln(2)}$ allows evaluation using natural logs.
• Since $2^x$ is one-to-one, the equation $2^x = a$ (with $a > 0$) has exactly one real solution.

Common Mistakes

• Forgetting to multiply the $-1$ term by $5$ when expanding $5(2^x - ​​1)$; this leads to an incorrect value for $2^x$.
• Sign errors when moving terms across the equation; write down each rearrangement carefully.
• Using the wrong log base or forgetting the change-of-base formula; remember $\log_2(a) = ​​\frac{\ln(a)}{\ln(2)}$.
• Rounding too early; keep the full calculator value until the final rounding to 3 significant figures.

Things to Be Careful About

• The original fraction is undefined when $2^x = ​​1$; our solution $2^x = ​​\frac{3}{2}$ does not cause this problem.
• The equation $2^x = a$ has a real solution only when $a > 0$; here $a = ​​\frac{3}{2} > 0$, so the logarithm step is valid.
• Correct to 3 significant figures means $0.585$, not $0.584$ or $0.5850$.
• The mark scheme also permits an iterative method; if using one, you must show the iteration and verify there are no other roots. The logarithm method gives the unique root directly.