Question 1
$e^{\ln(x)}$ is equivalent to
Not quite. A power of $e$ is always positive, so $e^{\ln(x)}$ can never be 0. Try $x=2$: $\ln(2)\approx0.693$ and $e^{0.693}\approx2$. In 2023, 5.9% of Queensland students chose (A).
Not quite. $e$ and $\ln$ undo each other, but what’s left is whatever was inside the log, not 1. $e^{\ln(x)}$ only equals 1 when $x=1$. At $x=2$, it equals 2. In 2023, 17.0% of Queensland students chose (B).
That’s it. Step through the hints below to see why. In 2023, 69.5% of Queensland students chose (C).
Not quite. $\dfrac{1}{x}$ is $e^{-\ln(x)}$, with a minus sign in the power. With no minus sign, $e^{\ln(x)}$ gives back $x$. At $x=2$, $\dfrac{1}{x}=0.5$ but $e^{\ln(2)}=2$. In 2023, 7.2% of Queensland students chose (D).
The dashed graph misses the black line here, so this option isn’t equivalent to $e^{\ln(x)}$.
At $x=1$ this option gives 1, and so does $e^{\ln(1)}=e^0$. One point can fool you. Move the guide and the dashed graph leaves the black line.
The two graphs are very close here, but the values in the readout are not equal. Move the guide to $x=2$ and the gap is easy to see.
The dashed line sits right on top of the black one at every $x$. Taking the log and then raising $e$ to that power gets you back where you started.
$\ln(x)$ and $e^x$ are inverse functions, so one undoes the other. $\ln(x)$ is the power you raise $e$ to in order to get $x$.
So raising $e$ to exactly that power gives $x$ back.
$$e^{\ln(x)}=x$$
Test $x=2$, where the options all give different values.
$\displaystyle \ln(2)\approx0.693$$\displaystyle e^{0.693\ldots}=2$
(A) gives 0, (B) gives 1, (C) gives 2 and (D) gives $\tfrac12$. Only (C) matches.
$\ln(x)$ only exists for $x\gt0$, so $e^{\ln(x)}=x$ holds for $x\gt0$. That is why the black line in the figure starts at an open circle at the origin.
Don’t test with $x=1$. There $e^{\ln(1)}=1$, and (B) and (D) also give 1, so that check can fool you.
$$e^{\ln(x)}=x$$
The exponential and the natural log are inverse functions. Raising $e$ to the power $\ln(x)$ undoes the log and gives back $x$, for every $x\gt0$.
How Queensland students answered in 2023
From the QCAA Mathematical Methods subject report for 2023. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2023 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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