Supplement: Logarithm Laws and Exponential Inverses

Formula reference. Require $b>0$, $b\ne1$, and every logarithm argument $>0$.

Rules

$\log_bx=y\iff b^y=x,\qquad \log_b(uv)=\log_bu+\log_bv,\qquad \log_b\!\left(\frac uv\right)=\log_bu-\log_bv.$

$\log_b(u^r)=r\log_bu,\qquad \ln(e^x)=x,\qquad e^{\ln x}=x\ (x>0),\qquad \log_bx=\frac{\ln x}{\ln b}.$

$\frac{d}{dx}\ln x=\frac1x\ (x>0),\qquad \frac{d}{dx}\ln|x|=\frac1x\ (x\ne0).$

Never split addition or subtraction: $\log_b(u+v)\ne\log_bu+\log_bv$ in general.

A logarithm answers an exponent question. The statement $\log_b(x)=y$ means exactly $b^y=x$, where $b>0$, $b\ne1$, and $x>0$. Thus the natural logarithm $\ln(x)$ is logarithm base $e$, not a new algebraic operation with unrelated rules. Keeping the inverse relationship visible is the safest way to interpret logarithms in calculus.

The input restriction $x>0$ is structural. In real-number calculus, $\ln(0)$ and $\ln(-3)$ are undefined because no real power of $e$ produces zero or a negative number. When a logarithm appears inside a function, solve its argument inequality before differentiating or graphing. A correct symbolic derivative cannot make an invalid domain valid.

A measurement story before the laws

Consider a sound-level comparison in which one signal has an intensity $I$ measured relative to a reference intensity $I_0$. A logarithmic scale reports a value proportional to $\log_{10}\!\left(\frac{I}{I_0}\right)$ because people often need to compare ratios that range over many powers of ten. A tenfold change is treated as one equal logarithmic step, while a hundredfold change is two steps. The logarithm is useful here precisely because multiplication in the original quantity becomes addition on the reported scale.

Suppose two independent effects multiply the intensity ratio: one factor comes from the source and one from transmission through a material. The reported log-scale changes add, because $\log_{10}(uv)=\log_{10}u+\log_{10}v$. This is not a typographical convenience; it lets a measurement system separate contributions that combine multiplicatively in the physical world. The conditions $u>0$ and $v>0$ reflect the fact that an intensity ratio must be positive.

We will return to this reporting story as each rule appears. The practical question is always: does the situation combine factors by multiplication or add quantities directly? Only the first structure creates a logarithm rule.

Derive the three core laws

For positive $u$ and $v$, $\log_b(uv)=\log_bu+\log_bv$. If $u=b^p$ and $v=b^q$, then $uv=b^{p+q}$, so its logarithm is $p+q$. Division works similarly: $\log_b\!\left(\frac uv\right)=\log_bu-\log_bv$. These rules translate multiplication and division inside a logarithm into addition and subtraction outside it.

The power law is $\log_b(u^r)=r\log_bu$ when the real expression is defined. It follows from $(b^p)^r=b^{pr}$. The law does not say $\log_b(u+v)=\log_bu+\log_bv$; addition has no matching logarithm rule. Try $u=v=1$ to see the contradiction: $\ln2$ is not $0+0$.

In the sound model, combining two intensity ratios multiplies them, so their logged values add. Adding two independent intensity contributions is a different physical operation and generally cannot be split with a log law. A calculator check is a useful guardrail, but the operation inside the logarithm is the deeper test. Ask “product, quotient, power, or sum?” before applying any transformation.

Read scale and solve equations

Logarithmic scales turn repeated multiplication into equal steps. A move from $10^2$ to $10^3$ changes a base-ten logarithm by one, even though the original quantity is multiplied by ten. This is why logs are useful for data spanning many orders of magnitude. The output grows slowly, but it remains defined only for positive inputs.

To solve $\ln(x-2)=3$, first preserve the domain condition $x-2>0$. Exponentiating both sides gives $x-2=e^3$, hence $x=2+e^3$. Substitution confirms that the logarithm receives a positive argument. Exponentiation is an inverse operation here, not a license to ignore restrictions.

The same inverse logic lets a technician recover an original ratio from a logarithmic report. If $\log_{10}(I/I_0)=4$, then $I/I_0=10^4$. The log scale did not lose the ratio; it encoded it as an exponent. Naming the base is essential, because the same numerical logarithm with a different base describes a different scale.

Prepare for differentiation

The derivative $\frac{d}{dx}\ln x=\frac1x$ is valid for $x>0$. With a positive inner function $g(x)$, the Chain Rule gives $\frac{d}{dx}\ln(g(x))=\frac{g'(x)}{g(x)}$. The denominator is not an accidental pattern: it reflects the logarithm’s domain and its sensitivity to relative rather than absolute change.

Logarithmic differentiation uses the power law to turn products and variable exponents into sums that are easier to differentiate. It is useful only after the original expression has a valid positive domain or an appropriate absolute-value form. Write the domain first, apply a law with its condition, then differentiate. That sequence guards against the most common mistakes.

Narrative challenge: interpret a logarithmic report

A sensor reports $L=\log_{10}\!\left(\frac{I}{I_0}\right)$. During one test the source increases the intensity ratio by a factor of $100$, while a filter reduces it by a factor of $10$. Before opening the solution, predict the net change in $L$ and explain why it is not appropriate to take the logarithm of “$100-10$.” Then state the condition on $I/I_0$ that allows the sensor formula to be used.

Show a solution path

The combined factor is $\frac{100}{10}=10$, so the change is $\log_{10}(10)=1$. Equivalently, the two log changes are $\log_{10}(100)-\log_{10}(10)=2-1=1$. The expression $100-10$ describes a difference, not the multiplicative combination modeled by the law. The ratio $I/I_0$ must be positive.