Mathematics 9709/32 — February/March 2017

Walkthrough

The equation has the unknown $x$ inside a logarithm. To undo the logarithm, exponentiate both sides: since $\ln$ and $e$ are inverse functions, $\ln(1+2^x)=2$ becomes $1+2^x=e^2$. This is the first key step and earns the B1 mark.

Next, isolate the term containing $x$ by subtracting $1$ from both sides. This gives $2^x=e^2-1$. Now the unknown is an exponent, so take logarithms. Using base $2$ directly gives $x=\log_2(e^2-1)$, or equivalently divide natural logs: $x=\frac{\ln(e^2-1)}{\ln(2)}$. This is the M1 method mark.

Finally, substitute $e^2-1=6.389056\ldots$ and compute the quotient. Keeping enough decimal places gives $x=2.6756\ldots$, which rounds to $2.676$ to three decimal places. This earns A1.

Key Takeaways

• Logarithmic equations can be solved by exponentiating both sides using the inverse relationship $\ln a=b \iff a=e^b$.
• When the unknown is in an exponent, taking logarithms (any consistent base) is the standard method.
• Always keep extra precision during calculation and round only at the final step.

Common Mistakes

• Forgetting to subtract $1$ after exponentiating: $2^x=e^2$ instead of $e^2-1$.
• Rounding intermediate values too early, which can change the third decimal place.
• Using the wrong base for logarithms or mixing bases in the same calculation.
• Giving an unsupported answer; the mark scheme requires showing the removal of logarithm and the method for solving $2^x=a$.

Things to Be Careful About

• The logarithm here is natural logarithm ($\ln$), so the inverse is $e$, not $10$.
• The answer must be given correct to $3$ decimal places, so write $2.676$, not $2.6756$ or $2.68$.
• Ensure the final value is positive; $e^2-1>0$, so the logarithm is defined.