eˣ, ln x, and Estimating Parameters from Logarithmic Graphs

What is the range of $f(x) = e^x$ for $x \in \mathbb{R}$, and hence what is the domain of $\ln x$?

• Range of $e^x$: $f(x) > 0$. Domain of $\ln x$: $x > 0$

  Correct. $e^x$ is strictly positive for every real $x$ and gets arbitrarily close to, but never reaches, $0$ — so its range excludes $0$ itself. Because $\ln x$ is the inverse of $e^x$, its domain is exactly this range: $x>0$.

• BRange of $e^x$: all real numbers. Domain of $\ln x$: all real numbers

  $e^x$ cannot be negative or zero for any real $x$ — there is no real $x$ solving $e^x = -1$ or $e^x=0$. The range is restricted to positive numbers, not the whole real line.

• CRange of $e^x$: $f(x) \geq 0$. Domain of $\ln x$: $x \geq 0$

  The restriction to non-negative values is right and the boundary is wrong: $e^x=0$ has no solution, so $0$ itself is never reached and must be excluded. The correct range is the strict inequality $f(x) > 0$.

• DRange of $e^x$: $f(x) > 0$. Domain of $\ln x$: all real numbers

  The range of $e^x$ is right; the domain of $\ln x$ has been left as if unrestricted instead of set equal to it. The domain of an inverse function is the range of the original — it doesn't stay at 'all real numbers' just because that happened to be true of $e^x$'s own domain.

Traps tested: Range assumed unrestricted · Boundary value included when limit never reached · Inverse domain range swap not applied

$y = e^{2x-6} + 1$. What is the equation of its asymptote, and its $y$-intercept to 3 s.f.?

• Asymptote $y=1$; $y$-intercept $\approx 1.00$

  Correct. As $x \to -\infty$, $e^{2x-6} \to 0$, so $y \to 1$: asymptote $y=1$. At $x=0$: $y = e^{-6}+1 \approx 0.00248 + 1 \approx 1.00$ (3 s.f.) — very close to the asymptote, since $e^{-6}$ is already small.

• BAsymptote $y=0$; $y$-intercept $\approx 0.00248$

  This finds the asymptote and intercept of $e^{2x-6}$ ALONE, before the $+1$ shift is applied. The $+1$ moves the entire curve — and its asymptote — up by 1 unit; both answers need that shift added.

• CAsymptote $y=-6$; $y$-intercept $\approx -5$

  This treats the $-6$ inside the exponent as if it were the vertical shift constant $c$ from $y=e^{ax+b}+c$. It isn't — the $-6$ is part of $b$, affecting the exponent's value, not shifting the curve vertically; the vertical shift is the standalone $+1$ outside the exponent entirely.

• DAsymptote $y=1$; $y$-intercept $\approx e^{-6} \approx 0.00248$

  The asymptote is right; the intercept has dropped the $+1$ partway through — it correctly finds $e^{-6}$ but never adds the $1$ back on to get the actual $y$-value at $x=0$.

Traps tested: Vertical shift constant dropped · Exponent constant confused with vertical shift

Solve $e^{2x-1} = 5$, giving your answer to 3 significant figures.

• $x \approx 1.30$

  Correct. Taking $\ln$ of both sides: $2x-1 = \ln 5$, so $x = \dfrac{\ln 5 + 1}{2} \approx \dfrac{1.609+1}{2} \approx 1.30$ (3 s.f.).

• B$x \approx 1.80$

  This divides by $2$ before adding the $1$: $\dfrac{\ln 5}{2}+1$, undoing the operations in the wrong order. The equation reads "double $x$, then subtract 1", so undoing it means adding 1 back FIRST, then dividing by 2 — the reverse order, exactly as with any other rearrangement.

• C$x \approx 0.305$

  This computes $\dfrac{\ln 5 - 1}{2}$ — the $-1$ from $2x-1$ has been subtracted again instead of added back on. Rearranging $2x-1=\ln 5$ for $x$ needs $+1$ on both sides first, not $-1$.

• D$x = 3$

  This solves $2x-1=5$ directly, as if the $\ln$ simply vanished rather than being applied to both sides. $\ln(e^{2x-1})$ collapses to $2x-1$, but $\ln(5)$ does NOT collapse to $5$ — the right-hand side genuinely needs the logarithm taking, it isn't already in the form the left side reduces to.

Traps tested: Order of undoing not reversed · Sign not flipped when undoing · Ln treated as removable without applying it

Solve $\ln(3x+2) = 1.5$, giving your answer to 3 significant figures.

• $x \approx 0.827$

  Correct. Applying $e^{(\ldots)}$ to both sides: $3x+2 = e^{1.5}$, so $x = \dfrac{e^{1.5}-2}{3} \approx \dfrac{4.482-2}{3} \approx 0.827$ (3 s.f.).

• B$x \approx -0.167$

  This treats $\ln(3x+2)=1.5$ as if it meant $3x+2=1.5$ directly, dropping the $\ln$ without undoing it properly. The equation needs $e^{(\ldots)}$ applied to BOTH sides — $e^{\ln(3x+2)}=3x+2$ on the left, $e^{1.5}$ on the right — not the logarithm simply erased.

• C$x \approx -0.506$

  This computes $\dfrac{e^{1.5}}{3}-2$ — dividing by 3 before subtracting 2, the wrong order for undoing "triple $x$, then add 2". Undo the addition first (subtract 2), then the multiplication (divide by 3).

• D$x \approx 1.49$

  This computes $\dfrac{e^{1.5}}{3} \approx 1.49$ and stops, having applied $e^{1.5}$ correctly but never subtracting the $2$ that was added to $3x$ in the first place — one genuine step of rearrangement is missing.

Traps tested: Ln treated as removable without applying it · Order of undoing not reversed · Rearrangement incomplete

A set of data pairs $(x, y)$ is believed to fit either $y = ax^n$ or $y = kb^x$. A graph of $\log_{10} y$ against $x$ (unlogged) turns out to be a straight line. Which model is this consistent with, and why?

• $y = kb^x$ — because logging only $y$ turns this form, and only this form, into a straight line in $x$

  Correct. Taking $\log_{10}$ of $y=kb^x$ gives $\log_{10}y = \log_{10}k + x\log_{10}b$, a straight line in the unlogged $x$ and logged $y$. $y=ax^n$ would need $\log_{10}x$ on the horizontal axis too — plotting it against unlogged $x$ instead would not straighten it.

• B$y = ax^n$ — because taking a log always straightens a curve, whichever form it started as

  Taking logs doesn't automatically straighten any curve — it straightens a SPECIFIC combination of axes for a SPECIFIC model form. $y=ax^n$ needs both log x and log y; plotted against unlogged x, it would still be a curve, not a line.

• CBoth models are equally consistent with this graph

  They aren't interchangeable — each model straightens under a specific, different pair of axes. A straight $\log_{10}y$-against-$x$ graph is diagnostic of $y=kb^x$ specifically; the same data plotted as $y=ax^n$ would only straighten on log-log axes, not these ones.

• DNeither — a straight line on any axes means $y$ is already linear in $x$

  The vertical axis here is $\log_{10}y$, not $y$ itself — a straight line in $(x, \log_{10}y)$ means $\log_{10}y$ is linear in $x$, which is a very different, and much richer, statement than $y$ itself being linear in $x$.

Traps tested: Log transform assumed to straighten any curve · Log linear model forms treated as interchangeable · Logged axis confused with unlogged variable

A graph of $\log_{10} y$ against $\log_{10} x$, for a relationship $y = ax^n$, is a straight line through $(0, 2)$ and $(2, 6)$. Find $n$ and $a$.

• $n = 2$, $a = 100$

  Correct. Gradient $= \dfrac{6-2}{2-0} = 2 = n$. The point $(0,2)$ already sits at $\log_{10}x=0$, so its $\log_{10}y$-value, $2$, IS the intercept: $\log_{10}a = 2$, so $a = 10^2 = 100$.

• B$n = 2$, $a = 2$

  The gradient (and hence $n$) is right; the intercept has been left as $\log_{10}a$ rather than converted to $a$ itself. $2$ is $\log_{10}a$, not $a$ — the un-logging step, $a = 10^2$, still has to happen.

• C$n = 100$, $a = 2$

  The two results have been swapped: the GRADIENT of a log-log graph is $n$, and the (un-logged) intercept is $a$ — not the other way round. Whichever number came from the RISE-over-RUN calculation is $n$; whichever came from a single point on the vertical axis needs un-logging to become $a$.

• D$n = -2$, $a = 100$

  The intercept conversion is right; the gradient has the wrong sign. $\dfrac{6-2}{2-0}=\dfrac{4}{2}=+2$, not $-2$ — the numerator and denominator (or their order) have been mixed up somewhere in the subtraction.

Traps tested: Intercept not converted from log form · Gradient and intercept roles swapped · Gradient sign error