Exponentials and logarithms

Explanation

• For $a>0$ with $a\neq1$, the exponential function $y=a^x$ has domain $\mathbb{R}$, range $y>0$, horizontal asymptote $y=0$ and intercept $(0,1)$.
• If $a>1$ the graph is increasing, while if $0<a<1$ it is decreasing; $a=1$ gives the constant function $y=1$.
• The natural exponential $e^x$ follows the same graph facts with base $e$, and transformations such as $Ae^{kx}+c$ scale, reflect and translate the basic graph.
• An exponential graph never reaches its horizontal asymptote; treating $a^0$ as $0$ instead of $1$ is a common error.

Explanation

• Differentiating gives $\dfrac{d}{dx}e^{kx}=ke^{kx}$, so the gradient of an exponential is a constant multiple of the function itself.
• For $Q=Q_0e^{kt}$, the rate satisfies $dQ/dt=kQ$: positive $k$ models growth and negative $k$ models decay.
• A tangent at $x=x_0$ uses the point $(x_0,e^{kx_0})$ and gradient $ke^{kx_0}$ in $y-y_0=m(x-x_0)$.
• This proportional-rate property is the mathematical reason for choosing an exponential model.
• Changing conditions can make its constant proportionality assumption unsuitable, so compare the model's rate with observed changes before accepting it as a valid model.

Explanation

• For $a>0$ with $a\neq1$, $\log_a x=y$ means exactly that $a^y=x$; a real logarithm requires $x>0$.
• The graph of $y=\log_a x$ is the reflection of $y=a^x$ in $y=x$, so its domain is $x>0$, range is $\mathbb{R}$ and vertical asymptote is $x=0$.
• Natural logarithms use base $e$, giving the inverse relations $\ln(e^x)=x$ for every real $x$ and $e^{\ln x}=x$ for $x>0$.
• The specification's inverse relationship does not make $\ln0$ defined: $e^x$ approaches zero but never equals it, so logarithm arguments must be strictly positive.

Explanation

• The laws convert between one logarithm and several: a product becomes a sum, a quotient a difference, and a power a coefficient, so $\ln\dfrac{x^3\sqrt y}{z^2}=3\ln x+\frac12\ln y-2\ln z$.
• Read backwards they collect an equation into a single logarithm, $\log_2(x+6)+\log_2(x-2)=3$ becoming $\log_2\big((x+6)(x-2)\big)=3$, which is then removed by writing the argument equal to $2^3$; “taking logs” of $y=ab^x$ gives $\log y=\log a+x\log b$, a straight line in $x$ whose intercept and gradient give $a$ and $b$.
• What the laws do not do: $\log(x+y)$ is not $\log x+\log y$, and $\log x\cdot\log y$ is not $\log(xy)$.
• Because a logarithm needs a positive argument, every solution is checked in the original equation and any that makes an argument zero or negative is rejected with that reason. $\log_a a=1$ and $\log_a1=0$.

Explanation

• For $a>0$, $a\neq1$ and $b>0$, taking logarithms gives $a^x=b\Rightarrow x=\log_a b=\ln b/\ln a$.
• If the exponent is linear, isolate the exponential or take logarithms first, then solve the resulting linear equation without rounding intermediate values.
• When both $a^x$ and $a^{-x}$ occur, substitute $u=a^x>0$, so $a^{-x}=1/u$, and solve the resulting algebraic equation before converting back.
• A positive-base exponential cannot equal zero or a negative number; after a substitution, reject non-positive values because $a^x$ is always positive.
• Substitute the final value into the original equation to verify it.

Explanation

• For $y=ax^n$, taking logarithms gives $\log y=\log a+n\log x$, so a plot of $\log y$ against $\log x$ has gradient $n$ and intercept $\log a$.
• For $y=kb^x$, $\log y=\log k+x\log b$, so a plot of $\log y$ against $x$ has gradient $\log b$ and intercept $\log k$.
• Read two well-separated points from the fitted line, calculate its gradient and intercept, then undo the chosen logarithm base to recover $a$, $k$ or $b$.
• The transformed coordinates and the logarithm base must be identified: confusing a log-log graph with a semi-log graph, or forgetting to exponentiate the intercept, gives incorrect parameters.

Explanation

• Repeated percentage change over whole periods uses $Q_n=Q_0(1+r)^n$, while continuous change at a rate proportional to the amount uses $Q=Q_0e^{kt}$.
• For decay $k<0$; the half-life in $Q=Q_0e^{-\lambda t}$ is $t_{1/2}=\ln2/\lambda$, independent of the starting amount.
• Use two observations to determine an unknown multiplier or exponent, preserve full precision, and interpret a calculated time or amount in the units and practical setting of the model.
• Unlimited exponential growth and a constant decay parameter may fail when resources, competition, dosage cycles or environmental conditions change; a carrying-capacity or piecewise model can be a refinement.