Change of base formula — proof: Exercises — Maths deep dive

Difficulty ladder used on this page:

L1 Recognition — spot the formula, plug in.
L2 Application — one clean computation.
L3 Analysis — pick a smart base, cancel structure.
L4 Synthesis — combine with the Laws of logarithms.
L5 Mastery — prove/solve where change of base is the only key that fits.

L1 — Recognition

Q1

Rewrite $lo g_{7} 20$ as a fraction of base-10 logs. (Do not evaluate.)

Recall Solution

WHAT: we translate old base $a = 7$ into new base $b = 10$ . Argument on top, old base on the bottom, new base on both: $lo g_{7} 20 = \frac{l o g _{10} 20}{l o g _{10} 7} .$ WHY that layout: the proof set $y = lo g_{7} 20$ , so $7^{y} = 20$ ; taking $lo g_{10}$ gives $y lo g_{10} 7 = lo g_{10} 20$ , and dividing puts the argument $20$ on top.

Q2

Rewrite $lo g_{3} 5$ using natural logs.

Recall Solution

WHAT: same move as Q1, but the new base is now $b = e$ (the base $ln$ measures in): $lo g_{3} 5 = \frac{l n 5}{l n 3} .$ WHY that layout: set $y = lo g_{3} 5$ , so by the definition $3^{y} = 5$ ; taking $ln$ of both sides gives $y ln 3 = ln 5$ , and dividing by $ln 3$ isolates $y$ — putting the argument $5$ on top and the old base $3$ on the bottom. WHY base $e$ here: the base of the new ruler is your free choice; $10$ and $e$ both give the identical number. We pick $e$ simply because the problem asks for natural logs, which is one of the two buttons a calculator provides.

L2 — Application

Q3

Compute $lo g_{2} 32$ using base 10. Give an exact integer.

Recall Solution

WHAT: apply the formula with old base $2$ , new base $10$ : $lo g_{2} 32 = \frac{l o g _{10} 32}{l o g _{10} 2} = \frac{1.50515}{0.30103} = 5.$ WHY those decimals: $lo g_{10} 32$ and $lo g_{10} 2$ are the two calculator numbers the translator asks for; neither is clean on its own, but their ratio must be exact because $32 = 2^{5}$ means "five twos multiplied", i.e. the answer to "how many 2s make 32?" is exactly $5$ . Check: $2^{5} = 32$ . ✓

Q4

Compute $lo g_{9} 40$ using $ln$ (3 d.p.).

Recall Solution

$lo g_{9} 40 = \frac{l n 40}{l n 9} = \frac{3.68888}{2.19722} \approx 1.679.$ Forecast-then-verify: $9^{1} = 9$ , $9^{2} = 81$ , and $40$ sits between, closer to the low end — so an answer just under halfway ( $\approx 1.68$ ) is sensible. ✓

Q5

Compute $lo g_{0.5} 8$ using base 10. (Note the base is between 0 and 1.)

Recall Solution

$lo g_{0.5} 8 = \frac{l o g _{10} 8}{l o g _{10} 0.5} = \frac{0.90309}{- 0.30103} = - 3.$ WHY negative: base $0.5 < 1$ means $lo g_{10} 0.5 < 0$ , so a positive numerator over a negative denominator gives a negative answer. Check: $0. 5^{- 3} = 2^{3} = 8$ . ✓ The formula handles a fractional base with no extra rules.

L3 — Analysis

Q6

Simplify $lo g_{8} x \cdot lo g_{x} 64$ for valid $x$ .

Recall Solution

WHAT: convert both to a common base $2$ (since $8 = 2^{3}$ , $64 = 2^{6}$ ). $\frac{l o g _{2} x}{l o g _{2} 8} \cdot \frac{l o g _{2} 64}{l o g _{2} x} = \frac{l o g _{2} x}{3} \cdot \frac{6}{l o g _{2} x} = \frac{6}{3} = 2.$ WHY common base: picking the shared base $2$ makes the $lo g_{2} x$ terms cancel, turning a product of two unknown logs into pure arithmetic.

Q7

Show that $lo g_{a} b \cdot lo g_{b} c \cdot lo g_{c} a = 1$ .

Recall Solution

WHAT: convert every factor to one fixed new base — call it $k$ (any valid base works; using a fresh letter avoids clashing with the $a, b, c$ already in the problem). Each $lo g_{p} q = \frac{lo g _{k} q}{lo g _{k} p}$ : $\frac{l o g _{k} b}{l o g _{k} a} \cdot \frac{l o g _{k} c}{l o g _{k} b} \cdot \frac{l o g _{k} a}{l o g _{k} c} .$ WHAT LOOKS LIKE: it's a chain where each numerator is the next denominator — a telescoping ring. Every $lo g_{k}$ term cancels, leaving $1$ .

Q8

Given $lo g_{2} 5 = m$ , express $lo g_{5} 8$ in terms of $m$ .

Recall Solution

WHAT: send $lo g_{5} 8$ back to base $2$ (the base $m$ lives in). $lo g_{5} 8 = \frac{l o g _{2} 8}{l o g _{2} 5} = \frac{3}{m} .$ $lo g_{2} 8 = 3$ because $2^{3} = 8$ , and $lo g_{2} 5 = m$ is given. So the answer is $\frac{3}{m}$ .

L4 — Synthesis

Q9

Solve $lo g_{x} 27 = \frac{3}{2}$ for $x > 0, x \neq = 1$ .

Recall Solution

WHAT: use the definition directly — $lo g_{x} 27 = \frac{3}{2}$ means $x^{3/2} = 27$ . Raise both sides to the power $\frac{2}{3}$ : $x = 2 7^{2/3} = (2 7^{1/3})^{2} = 3^{2} = 9.$ Check: $9^{3/2} = (9)^{3} = 3^{3} = 27$ . ✓

Q10

Solve $lo g_{4} x + lo g_{16} x = 3$ .

Recall Solution

WHAT: both bases are powers of $2$ , so convert to base $2$ . $\frac{l o g _{2} x}{l o g _{2} 4} + \frac{l o g _{2} x}{l o g _{2} 16} = \frac{l o g _{2} x}{2} + \frac{l o g _{2} x}{4} = 3.$ Let $u = lo g_{2} x$ : $\frac{u}{2} + \frac{u}{4} = \frac{3 u}{4} = 3 \Rightarrow u = 4$ . So $lo g_{2} x = 4 \Rightarrow x = 2^{4} = 16$ . Check: $lo g_{4} 16 + lo g_{16} 16 = 2 + 1 = 3$ . ✓

Q11

Solve $lo g_{3} x = lo g_{9} (2 x + 3)$ .

Recall Solution

WHAT: convert $lo g_{9}$ to base $3$ (since $9 = 3^{2}$ ): $lo g_{9} (2 x + 3) = \frac{l o g _{3} ( 2 x + 3 )}{l o g _{3} 9} = \frac{l o g _{3} ( 2 x + 3 )}{2} .$ Equation becomes $lo g_{3} x = \frac{1}{2} lo g_{3} (2 x + 3)$ , i.e. $2 lo g_{3} x = lo g_{3} (2 x + 3)$ , so $lo g_{3} x^{2} = lo g_{3} (2 x + 3)$ . Equal logs ⇒ equal arguments: $x^{2} = 2 x + 3 \Rightarrow x^{2} - 2 x - 3 = 0 \Rightarrow (x - 3) (x + 1) = 0$ . $x = 3$ or $x = - 1$ . Reject $x = - 1$ (a log needs a positive argument). So $x = 3$ . Check: $lo g_{3} 3 = 1$ ; $lo g_{9} 9 = 1$ . ✓

L5 — Mastery

Q12

Prove that for any valid bases, $\frac{1}{lo g _{2} N} + \frac{1}{lo g _{3} N} + \frac{1}{lo g _{5} N} = \frac{1}{lo g _{30} N}$ , where $N > 0, N \neq = 1$ .

Recall Solution

First, derive the reciprocal identity we need: $\frac{1}{lo g _{a} N} = lo g_{N} a$ . WHY it's true: apply change of base to $lo g_{N} a$ with new base $a$ : $lo g_{N} a = \frac{lo g _{a} a}{lo g _{a} N} = \frac{1}{lo g _{a} N}$ , since $lo g_{a} a = 1$ (because $a^{1} = a$ ). So flipping a log is the same as swapping its base and argument. Now the main proof. Using that identity termwise, the left side becomes $lo g_{N} 2 + lo g_{N} 3 + lo g_{N} 5.$ WHY this helps: now they share base $N$ , so the product law merges them: $lo g_{N} (2 \cdot 3 \cdot 5) = lo g_{N} 30 = \frac{1}{l o g _{30} N} . ■$ Change of base (via its reciprocal form) turned three awkward denominators into one clean sum.

Q13

Given $lo g_{12} 27 = a$ , express $lo g_{6} 16$ in terms of $a$ .

Recall Solution

WHAT: everything is built from primes $2$ and $3$ , so measure everything against base $2$ . Let $t = lo g_{2} 3$ (a single unknown ratio — naming it lets us do algebra instead of juggling logs). WHY convert $a$ first: we need one equation linking the given $a$ to $t$ , then a second expressing the target in $t$ . Change $lo g_{12} 27$ to base $2$ : $a = lo g_{12} 27 = \frac{l o g _{2} 27}{l o g _{2} 12} = \frac{3 l o g _{2} 3}{2 + l o g _{2} 3} = \frac{3 t}{2 + t} .$ Here $lo g_{2} 27 = lo g_{2} 3^{3} = 3 lo g_{2} 3 = 3 t$ (power law) and $lo g_{2} 12 = lo g_{2} (4 \cdot 3) = 2 + lo g_{2} 3 = 2 + t$ (product law, $lo g_{2} 4 = 2$ ). Solve for $t$ (clear the fraction so $t$ is isolated): $a (2 + t) = 3 t \Rightarrow 2 a = 3 t - a t = t (3 - a) \Rightarrow t = \frac{2 a}{3 - a}$ . Now the target, also in base $2$ : $lo g_{6} 16 = \frac{l o g _{2} 16}{l o g _{2} 6} = \frac{4}{1 + l o g _{2} 3} = \frac{4}{1 + t} .$ ( $lo g_{2} 16 = 4$ ; $lo g_{2} 6 = lo g_{2} (2 \cdot 3) = 1 + t$ .) Substitute $t$ and simplify the compound fraction (multiply top and bottom by $3 - a$ ): $lo g_{6} 16 = \frac{4}{1 + \frac{2 a}{3 - a}} = \frac{4 ( 3 - a )}{( 3 - a ) + 2 a} = \frac{4 ( 3 - a )}{3 + a} .$ Check numerically: $a = lo g_{12} 27 \approx 1.3263$ ; formula gives $\frac{4 ( 3 - 1.3263 )}{3 + 1.3263} \approx 1.547$ , and directly $lo g_{6} 16 = \frac{l n 16}{l n 6} \approx 1.547$ . ✓

Q14

Solve $x^{l o g_{10} x} = 100 x$ for $x > 0$ .

Recall Solution

WHAT: take $lo g_{10}$ of both sides to bring the exponent down — that's exactly what a log is for. $lo g_{10} (x^{l o g_{10} x}) = lo g_{10} (100 x) .$ Left: power law gives $(lo g_{10} x)^{2}$ . Right: $lo g_{10} 100 + lo g_{10} x = 2 + lo g_{10} x$ . Let $u = lo g_{10} x$ : $u^{2} = 2 + u \Rightarrow u^{2} - u - 2 = 0 \Rightarrow (u - 2) (u + 1) = 0.$ $u = 2 \Rightarrow x = 1 0^{2} = 100$ , or $u = - 1 \Rightarrow x = 1 0^{- 1} = 0.1$ . Both give positive $x$ , so both valid. Check $x = 100$ : LHS $= 10 0^{2} = 1 0^{4}$ , RHS $= 100 \cdot 100 = 1 0^{4}$ . ✓ Check $x = 0.1$ : LHS $= 0. 1^{- 1} = 10$ , RHS $= 100 \cdot 0.1 = 10$ . ✓

Active recall

Connections

3.2.09 Change of base formula — proof — the parent these drills train.
Definition of a logarithm — Q9 solves straight from it.
Laws of logarithms — the power/product laws drive Q10–Q14.
Natural logarithm ln — the free-choice new base in Q2, Q4, Q13.
Exponential functions — the log↔exponent switch underlying every solution.
Solving exponential equations — Q14's "log both sides" technique.