Laws of logarithms — product, quotient, power rules — proofs: Question bank

True or false — justify

Each line: decide true or false, then give the one-sentence reason before revealing.

lo g_{b} (M N) = lo g_{b} M + lo g_{b} N

for all

M, N > 0

True. This is the product rule — same-base multiplication adds exponents, and logs read off those exponents.

lo g_{b} (M + N) = lo g_{b} M + lo g_{b} N

False. Logs turn products into sums, never sums into sums;

lo g_{b} (M + N)

has no simplification. Test:

lo g_{10} (1 + 1) = 0.301

but

lo g_{10} 1 + lo g_{10} 1 = 0

lo g_{b} (M - N) = lo g_{b} M - lo g_{b} N

False. Subtraction of logs comes from division inside, i.e.

lo g_{b} (M / N)

, not from subtraction inside; the two are unrelated.

(lo g_{b} M)^{2} = 2 lo g_{b} M

False. The power rule only brings down an exponent that sits on the argument $M$ ; here the whole log is squared, so

(lo g_{b} M)^{2} = lo g_{b} M \cdot lo g_{b} M

— no simplification.

lo g_{b} (M^{2}) = 2 lo g_{b} M

for

M > 0

True. The exponent is on the argument, so the power rule applies and pulls the

2

out front.

\frac{lo g M}{lo g N} = lo g_{b} M - lo g_{b} N

False. A ratio of two logs is the change-of-base result

lo g_{N} M

(see Change of base formula), not a subtraction.

lo g_{b} (1/ M) = - lo g_{b} M

for

M > 0

True. Take the power rule with exponent

k = - 1

lo g_{b} (M^{- 1}) = - 1 \cdot lo g_{b} M

lo g_{b} (b^{k}) = k

for every real

k

True. By definition the log asks "what power gives

b^{k}

?" — and the answer is literally

k

, even for negative or fractional

k

lo g_{b} 1 = 0

for every valid base

b

True. Any nonzero base to the power

0

1

, so the exponent needed to reach

1

is always

0

lo g_{b} b = 1

for every valid base

b

True. You raise

b

to the power

1

to get

b

, so the log is

1

The power rule

lo g_{b} (M^{k}) = k lo g_{b} M

only works for whole-number

k

False. It holds for any real

k

— including negatives, fractions, and irrationals — because

(b^{p})^{k} = b^{p k}

is true for all real

k

lo g_{b} (M N) = (lo g_{b} M) (lo g_{b} N)

False. The product inside becomes a sum of logs, not a product of logs; multiplying the logs is a different, wrong operation.

Spot the error

Each line states a "proof" or step that is wrong. Name the exact broken move.

lo g_{2} (8 \times 4) = lo g_{2} 8 \times lo g_{2} 4 = 3 \times 2 = 6

The product rule adds logs, it does not multiply them: the answer is

3 + 2 = 5

, and indeed

lo g_{2} 32 = 5

lo g_{3} (x^{2}) = (lo g_{3} x)^{2}

because the

2

just moves."

The

2

is an exponent on the argument, so it comes out as a coefficient:

lo g_{3} (x^{2}) = 2 lo g_{3} x

, not a squared log.

"Solve

lo g_{5} x + lo g_{5} (x - 4) = 1

: get

x = 5

x = - 1

, so both are answers."

x = - 1

makes

lo g_{5} x

undefined (argument must be

> 0

), so it must be discarded — only

x = 5

survives.

lo g_{b} 0 = 0

since anything to the

0

is... something."

lo g_{b} 0

is undefined: no real power of a positive base ever reaches

0

, so the question has no answer.

lo g_{b} (M / N) = \frac{lo g _{b} M}{lo g _{b} N}

— division inside becomes division outside."

Division inside becomes subtraction:

lo g_{b} (M / N) = lo g_{b} M - lo g_{b} N

; the ratio-of-logs form is change of base, a separate identity.

3 lo g_{2} x = lo g_{2} 3 x

— bring the

3

inside as a multiplier."

The coefficient becomes an exponent, not a factor:

3 lo g_{2} x = lo g_{2} x^{3}

- lo g_{b} M = lo g_{b} (- M)

— the minus just moves onto the argument."

A leading minus means reciprocal inside:

- lo g_{b} M = lo g_{b} (1/ M)

; and

lo g_{b} (- M)

is undefined for

M > 0

anyway.

lo g_{2} 48 - lo g_{2} 3 = lo g_{2} 48 \div lo g_{2} 3

Subtraction of logs collapses the arguments by division inside one log:

lo g_{2} (48/3) = lo g_{2} 16 = 4

Why questions

lo g_{b} x

requires

x > 0

— why?

A positive base raised to any real power is always positive, so no exponent can ever produce a zero or negative result to take the log of.

The base must satisfy

b \neq = 1

— why?

1

raised to any power is always

1

, so

lo g_{1} x

could never distinguish or reach any other number; the log would be undefined.

The base must satisfy

b > 0

— why?

Negative bases give non-real results for fractional powers (e.g.

(- 2)^{1/2}

), so the exponential

b^{y}

wouldn't be a well-behaved real function to invert.

Why does multiplying numbers turn into adding their logs?

Because logs are exponents and same-base powers multiply by adding exponents (

b^{p} b^{q} = b^{p + q}

); the log just reads off that sum.

Why does the power rule pull the exponent out as a multiplier?

A power of a power multiplies exponents (

(b^{p})^{k} = b^{p k}

), so

k

copies of the log's exponent become

k

times it.

Why is the quotient rule not really a separate law?

It is the product rule combined with

lo g_{b} (1/ N) = - lo g_{b} N

: dividing by

N

is multiplying by

N^{- 1}

Why must you check candidate solutions after using log rules to solve an equation?

Combining logs can hide domain restrictions; a solution valid for the merged equation may make an original argument

\leq 0

and be invalid.

Why can

lo g_{b} (b^{k}) = k

hold even when

k

is negative?

The definition places no sign restriction on the exponent itself — only on the argument

b^{k}

, which stays positive for any real

k

Edge cases

What is

lo g_{b} (M^{k})

when

k = 0

(and

M > 0

M^{0} = 1

, so

lo g_{b} 1 = 0

; consistently, the power rule gives

0 \cdot lo g_{b} M = 0

What happens to the product rule if one factor equals

1

lo g_{b} (M \cdot 1) = lo g_{b} M + lo g_{b} 1 = lo g_{b} M + 0

, so multiplying by

1

correctly changes nothing.

Can you take

lo g_{b}

of a very small positive number like

0.001

Yes — small positive arguments are fine and give large negative logs (e.g.

lo g_{10} 0.001 = - 3

); only

0

and negatives are forbidden.

lo g_{b} (M / M)

defined, and what is it?

Yes, for

M > 0

lo g_{b} (M / M) = lo g_{b} M - lo g_{b} M = lo g_{b} 1 = 0

Does the power rule let

M

be negative if

k

is even, since

M^{k} > 0

No — the rule requires

M > 0

from the start; although

M^{k}

may be positive,

lo g_{b} M

on the right side would itself be undefined.

What is

lo g_{b} b^{0}

It equals

0

, since

b^{0} = 1

and the log of

1

is always

0

— a clean boundary check on

lo g_{b} (b^{k}) = k

x \to 0^{+}

, what does

lo g_{b} x

do (for

b > 1

It heads to

- \infty

: you need ever-more-negative exponents to reach ever-smaller positive numbers, so there is no log of

0

itself (see Graphs of logarithmic functions).

Connections

Laws of logarithms — product, quotient, power rules — proofs — the proofs these traps stress-test.
Exponentials — laws of indices — the index laws every trap ultimately traces back to.
Change of base formula — resolves the "ratio of logs vs. subtraction" confusion.
Solving exponential equations with logs — where the domain-checking edge cases bite.
Natural logarithm ln and e — the same traps in base $e$ .
Graphs of logarithmic functions — the picture behind the $x \to 0^{+}$ and domain edge cases.