Exercises — Common log (log₁₀) and natural log (ln x)

WHAT we ask: "10 to what power gives $100000$?" Count the zeros: $100000=10^5$ (five factors of ten). Why? Each zero is one more ×10. Read the exponent: ${\log }_{10}(10^5)=5$. $\boxed{{\log }_{10}(100000)=5}$

WHAT we ask: "$e$ to what power gives $e^7$?" The base of $\ln$ is $e$, and the number is already written as a power of $e$. So $\ln (e^7)$ asks "which power?" and the number tells us: it's $7$. $\boxed{\ln (e^7)=7}$ This is the "log and exp cancel" fact: $\ln (e^u)=u$ for any $u$.

Rewrite as a power of ten first. Two decimal places means two factors of $\frac{1}{10}$: $0.01=\frac{1}{100}=\frac{1}{10^2}=10^{-2}.$ Why the minus sign? Dividing by $10$ is the same as multiplying by $10^{-1}$; do it twice → $10^{-2}$. Read the exponent: ${\log }_{10}(10^{-2})=-2$. $\boxed{{\log }_{10}(0.01)=-2}$

WHY take a log? The unknown $x$ is stuck in the exponent. A log is the exact tool that pulls exponents down (product/power law), so we apply ${\log }_{10}$ to both sides. ${\log }_{10}(10^x)={\log }_{10}250\Rightarrow x={\log }_{10}250.$ $x={\log }_{10}250\approx \mathrm{2.3979.}$ Sanity check: $250$ is between $100=10^2$ and $1000=10^3$, so $x$ should be between $2$ and $3$. ✅ $\boxed{x\approx 2.3979}$

WHY $\ln$ here and not ${\log }_{10}$? The base of the exponential is $e$, so $\ln$ (base $e$) cancels it cleanly with no extra constant. Apply $\ln$ to both sides: $\ln (e^{5x})=\ln 42\Rightarrow 5x=\ln 42.$ Why did the LHS become $5x$? Because $\ln (e^u)=u$; here $u=5x$. Isolate $x$ by dividing by $5$: $x=\frac{\ln 42}{5}\approx \frac{3.7377}{5}\approx \mathrm{0.7475.}$ $\boxed{x\approx 0.7475}$

The base is $3$, and our calculator only has base-$10$ and base-$e$ buttons. Take $\ln$ of both sides and let the power law bring $x$ down: $\ln (3^x)=\ln 90\Rightarrow x\ln 3=\ln 90.$ Divide by $\ln 3$ (a constant number, $\approx 1.0986$): $x=\frac{\ln 90}{\ln 3}\approx \frac{4.4998}{1.0986}\approx \mathrm{4.0959.}$ This is exactly the change-of-base formula ${\log }_{3}90=\frac{\ln 90}{\ln 3}$ in action. $\boxed{x\approx 4.0959}$

Combine with product then quotient laws. Sums become products, the subtraction becomes a divide: ${\log }_{10}8+{\log }_{10}25-{\log }_{10}2={\log }_{10}\left(\frac{8\times 25}{2}\right).$ Simplify inside: $\frac{8\times 25}{2}=\frac{200}{2}=100$. $={\log }_{10}(100)={\log }_{10}(10^2)=2.$ $\boxed{2}$ Notice we never touched a calculator — the pieces were designed to collapse to a power of ten.

Combine the left side with the product law: ${\log }_{10}(x(x-3))=1.$ Undo the log using the definition (${\log }_{10}A=1⟺A=10^1$): $x(x-3)=10\Rightarrow x^2-3x-10=0.$ Factor: $(x-5)(x+2)=0$, so $x=5$ or $x=-2$. Check the domain — CRUCIAL for logs. We need both $x>0$ and $x-3>0$, i.e. $x>3$.

• $x=5$: gives ${\log }_{10}5+{\log }_{10}2={\log }_{10}10=1$ ✅ — valid.
• $x=-2$: makes ${\log }_{10}(-2)$ undefined ❌ — reject. $\boxed{x=5}$

Clever move: $5=\frac{10}{2}$. Use the quotient law so the $10$ gives an exact $1$: ${\log }_{10}5={\log }_{10}\left(\frac{10}{2}\right)={\log }_{10}10-{\log }_{10}2=1-\mathrm{0.3010.}$ $\boxed{{\log }_{10}5\approx 0.6990}$ Why this works: anything you can build out of $2$s and $10$s, you can log by hand once you know ${\log }_{10}2$.

Set up the "half" condition. We want $N(t)=\frac{1}{2}{N}_0$: $N_0{e}^{-0.045t}=\frac{1}{2}{N}_0.$ Divide by $N_0$ (the starting amount, non-zero) — this is why half-life doesn't depend on how much you start with: $e^{-0.045t}=\frac{1}{2}.$ Take $\ln$ (base $e$ matches the $e$ in the model, so it cancels cleanly): $-0.045t=\ln \left(\frac{1}{2}\right)=-\ln 2.$ Why is $\ln (1/2)=-\ln 2$? Because $\frac{1}{2}=2^{-1}$ and the power law slides the $-1$ out front. Solve for $t$ (two negatives cancel, so $t>0$ as a time should be): $t=\frac{\ln 2}{0.045}\approx \frac{0.6931}{0.045}\approx 15.40\text{ years}.$ $\boxed{t\approx 15.40\text{ years}}$

Turn each magnitude back into an intensity ratio using the definition of log: $\frac{I_A}{I_0}=10^{6.2},\frac{I_B}{I_0}=10^{4.7}.$ Divide A by B so the common reference $I_0$ cancels: $\frac{I_A}{I_B}=\frac{10^{6.2}}{10^{4.7}}=10^{6.2-4.7}=10^{1.5}.$ Why subtract exponents? Dividing same-base powers subtracts exponents — the exponential form of the quotient law. $10^{1.5}\approx \mathrm{31.62.}$ $\boxed{\text{A is about }31.6\text{ times more intense than B.}}$ Lesson: a difference of just $1.5$ on a log scale is a factor of $\approx 31$ in the real quantity — that's the whole point of log scales.

Start from change of base (from the parent note), writing both sides in natural logs: ${\log }_{b}a=\frac{\ln a}{\ln b},{\log }_{a}b=\frac{\ln b}{\ln a}.$ Multiply them: ${\log }_{b}a\cdot {\log }_{a}b=\frac{\ln a}{\ln b}\cdot \frac{\ln b}{\ln a}=1.$ Why does everything cancel? The numerator of one is the denominator of the other — the fractions are literal reciprocals. Since the product is $1$, each is the reciprocal of the other: $\boxed{{\log }_{b}a=\frac{1}{{\log }_{a}b}.}$ Apply with $b=10,a=e$: ${\log }_{10}e=\frac{1}{{\log }_{e}10}=\frac{1}{\ln 10}\approx \mathrm{0.4343.}$ This is exactly the "Forecast-then-Verify" fact from the parent note.

Take $\ln$ of both sides so the two different bases both surrender their exponents: $\ln (2^x)=\ln (3^{x-1})\Rightarrow x\ln 2=(x-1)\ln 3.$ Expand the right side and gather all $x$ terms on one side: $x\ln 2=x\ln 3-\ln 3\Rightarrow x\ln 2-x\ln 3=-\ln 3.$ Factor out $x$: $x(\ln 2-\ln 3)=-\ln 3\Rightarrow x=\frac{-\ln 3}{\ln 2-\ln 3}=\frac{\ln 3}{\ln 3-\ln 2}.$ Why flip both signs? Multiplying top and bottom by $-1$ keeps the value but makes the denominator $\ln 3-\ln 2>0$, which reads more cleanly. Compute: $\frac{\ln 3}{\ln 3-\ln 2}=\frac{1.0986}{1.0986-0.6931}=\frac{1.0986}{0.4055}\approx \mathrm{2.7095.}$ $\boxed{x=\frac{\ln 3}{\ln 3-\ln 2}\approx 2.7095}$

Work from the outside in — but read the gates from the inside out.

• Inner gate: for ${\log }_{10}x$ to exist we need $x>0$.
• Outer gate: the outer ${\log }_{10}$ needs its input ${\log }_{10}x$ to be strictly positive: ${\log }_{10}x>0.$ Solve the outer condition. ${\log }_{10}x>0$ means "$x$ is a positive power of ten," which happens exactly when $x>1$ (since ${\log }_{10}1=0$ and logs grow past that). Combine: $x>1$ already forces $x>0$, so the binding condition is: $\boxed{x>1.}$ Check the boundary: at $x=1$, ${\log }_{10}1=0$, and ${\log }_{10}(0)$ is undefined — confirming we need strictly greater than $1$, not $\ge$.