Exercises — Solving exponential equations using logarithms

WHAT: the unknown $x$ sits in the exponent, so ordinary algebra can't free it → we need a log. Step 1 (Log both sides): $\log (3^x)=\log 20$. Step 2 (Drop): power law $\log (m^k)=k\log m$ turns the exponent into a multiplier: $x\log 3=\log 20.$ Step 3 (Solve): $\log 3$ is now just a fixed number, so divide: $x=\frac{\log 20}{\log 3}=\frac{1.3010}{0.4771}\approx \mathrm{2.727.}$ Sanity check: $3^2=9$ and $3^3=27$, and $20$ lies between → $x$ between $2$ and $3$. ✓

Because the base is 10, taking ${\log }_{10}$ is the inverse operation directly (see Inverse Functions): $\log (10^x)=\log 4500\Rightarrow x=\log 4500\approx \mathrm{3.653.}$ Why so quick? ${\log }_{10}(10^x)=x$ by definition of an inverse — the log and the base-10 power cancel with nothing left over. Check: $10^3=1000$, $10^4=10000$, $4500$ is between → $x$ between $3$ and $4$. ✓

This is a degenerate / edge case — do NOT reach for a calculator yet. $x\log 2=\log 1.$ The key fact: $\log 1=0$ because $10^0=1$ (any base to the power $0$ is $1$). So $x\log 2=0\Rightarrow x=\frac{0}{\log 2}=0.$ Why it must be $0$: $b^0=1$ for every base $b\ne 0$. The log method still works but the answer is exactly $0$, no rounding needed. ✓

Isolate first: divide by $4$ → $5^x=25$. Spot it: $25=5^2$, so exact answer $x=2$ without any log! But by method: $x\log 5=\log 25\Rightarrow x=\frac{\log 25}{\log 5}=\frac{1.3979}{0.6990}=\mathrm{2.000.}$ Why divide by 4 before logging? The $4$ multiplies the whole power; if you logged too early you'd get $\log 4+x\log 5$ (product law) and a messier line. Isolating keeps it clean. ✓

Isolate the power: subtract $3$ → $7^x=47$. (You subtract, not divide, because $3$ is added, not multiplied.) $x\log 7=\log 47\Rightarrow x=\frac{\log 47}{\log 7}=\frac{1.6721}{0.8451}\approx \mathrm{1.979.}$ Check: $7^2=49\approx 47$, so $x$ just under $2$. ✓

Isolate: divide by $200$ → $(0.5{)}^x=0.125$. $x\log (0.5)=\log (0.125).$ Here's the twist: $\log (0.5)$ is negative (since $0.5<1$, and $\log$ of a number below 1 is negative). Both sides negative: $x=\frac{\log 0.125}{\log 0.5}=\frac{-0.9031}{-0.3010}=\mathrm{3.000.}$ Why still positive $x$? Two negatives divide to a positive. Note $0.5^3=0.125$ exactly. This is a decay base $<1$ — see Exponential Growth and Decay. ✓

Log both sides: $\log (2^{3x+1})=\log 30$. Drop the WHOLE exponent as a bracket: $(3x+1)\log 2=\log 30.$ Solve the linear part: $3x+1=\frac{\log 30}{\log 2}=\frac{1.4771}{0.3010}=4.9069,$ $3x=3.9069\Rightarrow x=\mathrm{1.302.}$ Why bracket it? The exponent is one connected object $3x+1$; the power law multiplies all of it by $\log 2$. Forget the bracket and you'd wrongly get $3x\log 2+1$. ✓

$\log (5^{2-x})=\log 12\Rightarrow (2-x)\log 5=\log 12.$ $2-x=\frac{\log 12}{\log 5}=\frac{1.0792}{0.6990}=1.5440,$ $-x=-0.4560\Rightarrow x=\mathrm{0.4560.}$ Watch the sign: the $-x$ means the answer is $x=2-1.5440$. Careful subtraction, not $x=1.5440-2$. ✓

WHY logs, not base-matching: $3$ and $5$ share no common base, so the "match the exponents" trick fails. Log is the only route. Log both sides: $2x\log 3=(x+1)\log 5.$ Expand (both brackets must open): $2x\log 3=x\log 5+\log 5.$ Collect $x$-terms on one side: $2x\log 3-x\log 5=\log 5,$ $x(2\log 3-\log 5)=\log 5.$ Solve: $x=\frac{\log 5}{2\log 3-\log 5}=\frac{0.6990}{2(0.4771)-0.6990}=\frac{0.6990}{0.2552}\approx \mathrm{2.739.}$ Why factor out $x$? Once every $x$ is on one side, $x$ is a common factor — pull it out and divide, exactly like solving $ax=b$. ✓

Base-matching beats logs here: $4=2^2$ and $8=2^3$, both powers of $2$. $4^x=(2^2{)}^x=2^{2x},8^{x-1}=(2^3{)}^{x-1}=2^{3(x-1)}=2^{3x-3}.$ Equation becomes $2^{2x}=2^{3x-3}$. If bases are equal, exponents are equal: $2x=3x-3\Rightarrow 3=x\Rightarrow x=3.$ Why exact? Rewriting to a common base avoids $\log$ entirely — no rounding at all. Use it whenever bases are powers of a common number. ✓

Spot the disguise: let $u=2^x$. Then $2^{2x}=(2^x{)}^2=u^2$. The equation becomes a plain quadratic: $u^2-5u+4=0\Rightarrow (u-1)(u-4)=0.$ So $u=1$ or $u=4$, i.e. $2^x=1$ or $2^x=4$.

• $2^x=1\Rightarrow x=0$ (since $2^0=1$).
• $2^x=4\Rightarrow x=2$ (since $2^2=4$). Both are valid — $2^x$ is always positive, and both $u$-values ($1$ and $4$) are positive, so neither is rejected. Answers: $x=0$ and $x=2$. Why substitute? The equation looks exponential but is structurally quadratic in $2^x$; naming that block $u$ reveals the familiar factorisation. ✓

Let $u=3^x$ (so $u>0$ always). Then $3^{2x}=u^2$: $u^2-4u-5=0\Rightarrow (u-5)(u+1)=0\Rightarrow u=5\text{ or }u=-1.$ Reject $u=-1$: $3^x=-1$ has no solution because $3^x$ is positive for every real $x$ (an exponential never dips to or below zero). This is the crucial edge case. Keep $u=5$: $3^x=5\Rightarrow x=\frac{\log 5}{\log 3}=\frac{0.6990}{0.4771}\approx \mathrm{1.465.}$ Answer: $x\approx 1.465$ only. ✓

Set up the equation (doubling means value $=1000$): $500\cdot (1.06{)}^t=1000.$ Isolate the power: divide by $500$ → $(1.06{)}^t=2$. Log and drop: $t\log 1.06=\log 2\Rightarrow t=\frac{\log 2}{\log 1.06}=\frac{0.3010}{0.02531}\approx \mathrm{11.90.}$ Interpret: at $t=11.90$ years it hits exactly £1000; since interest is added yearly, it first exceeds £1000 after $t=12$ whole years. Why round UP here? The question asks when it first exceeds, and $11.90$ years isn't a whole year — you must wait to year $12$. See Exponential Growth and Decay. ✓

Set up: "$10\%$ remains" means $N=0.1N_0$: $N_0(0.5{)}^{t/8}=0.1N_0.$ Cancel $N_0$ (it's on both sides, and $N_0\ne 0$): $(0.5{)}^{t/8}=0.1$. Log both sides, drop the bracketed exponent $t/8$: $\frac{t}{8}\log 0.5=\log \mathrm{0.1.}$ $\frac{t}{8}=\frac{\log 0.1}{\log 0.5}=\frac{-1}{-0.3010}=\mathrm{3.3219.}$ $t=8\times 3.3219\approx 26.58\text{ hours}.$ Sanity check with half-lives: after $8$h → $50\%$, $16$h → $25\%$, $24$h → $12.5\%$, $32$h → $6.25\%$. $10\%$ sits between $24$h and $32$h → our $26.58$h is right. ✓

Set them equal: $1000\cdot (1.03{)}^t=600\cdot (1.05{)}^t.$ Group the powers — divide both sides by $(1.03{)}^t$ and by $600$: $\frac{1000}{600}=\frac{(1.05{)}^t}{(1.03{)}^t}={\left(\frac{1.05}{1.03}\right)}^t.$ So ${\left(\frac{1.05}{1.03}\right)}^t=\frac{5}{3}=1.6667$, with $\frac{1.05}{1.03}=1.019417$. Log and drop: $t\log (1.019417)=\log (1.6667),$ $t=\frac{\log 1.6667}{\log 1.019417}=\frac{0.2218}{0.008353}\approx \mathrm{26.55.}$ Interpret: the slower-starting-but-faster-growing B catches A at about $t\approx 26.6$ years. (Cross-check graphically with Solving Equations Graphically — the two curves intersect there.) ✓

1. Is the power alone? If not, undo the blocker (÷ for $\times$, − for $+$).
2. Do the bases match as powers of one number? If yes → equate exponents (exact, no logs).
3. Is it secretly a quadratic in $b^x$? If yes → substitute $u=b^x$, solve, reject $u\le 0$.
4. Otherwise → log both sides, drop the whole bracketed exponent, solve linearly.
5. Round toward the question's real meaning, not blindly.