Logarithm — definition as inverse of exponential: Question bank

True or false — justify

A true/false is only worth points if you can say why. Give the reason, not the verdict.

lo g_{2} 8 = 3

because

2^{3} = 8

True. The log asks "2 to what power is 8?" and

2 \times 2 \times 2 = 8

, so the exponent is

3

lo g_{b} 1 = 0

for every valid base

b

True. For any valid base (

b > 0, b \neq = 1

) we have

b^{0} = 1

, so the exponent producing

1

is always

0

— the same for every such base.

lo g_{b} b = 1

for every valid base

b

True.

b^{1} = b

, so the exponent that turns

b

into

b

1

. This is why every

lo g_{b} x

graph passes through

(b, 1)

lo g_{2} (- 8) = - 3

because

2^{- 3}

has a minus in it.

False.

2^{- 3} = \frac{1}{8}

, a positive number. No real power of a positive base is ever negative, so

lo g_{2} (- 8)

is undefined. The minus in an exponent shrinks the value, it never flips its sign.

lo g_{10} 0 = 0

False. It is undefined. To get

0

you would need

1 0^{x} = 0

, but

1 0^{x}

shrinks toward

0

without ever reaching it as

x \to - \infty

. There is no exponent that lands exactly on

0

lo g_{3} (x + y) = lo g_{3} x + lo g_{3} y

False. Logs turn multiplication into addition (

lo g_{b} (x y) = lo g_{b} x + lo g_{b} y

), not addition into addition. Test:

lo g_{3} (3 + 3) = lo g_{3} 6 \approx 1.63

, but

lo g_{3} 3 + lo g_{3} 3 = 2

. See the laws of logarithms.

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

holds for negative

x

too.

True, for every real

x

. The exponent that produces

b^{x}

is literally

x

whether

x

is positive, zero or negative — the cancellation identity has no sign restriction.

b^{l o g_{b} y} = y

holds for every real number

y

False. It holds only for

y > 0

. The expression

lo g_{b} y

does not exist for

y \leq 0

, so you cannot even write the left side otherwise.

For

0 < b < 1

, the function

b^{x}

still has an inverse.

True.

b^{x}

is strictly decreasing when

0 < b < 1

, and strictly decreasing functions are still one-to-one, so an inverse

lo g_{b}

exists. Only the base

b = 1

(constant) fails.

lo g_{5} 5^{2.3} = (lo g_{5} 5)^{2.3}

False. Left side

= 2.3

(the exponent is right there). Right side

= 1^{2.3} = 1

since

lo g_{5} 5 = 1

. The brackets sit in completely different places, giving different meanings.

Spot the error

Each line contains a plausible-looking mistake. Find it and repair the reasoning.

lo g_{2} 8 = 4

because

8

has four factors of... something."

Go back to the definition:

lo g_{2} 8 = x

means

2^{x} = 8

. Count the doublings:

2^{1} = 2, 2^{2} = 4, 2^{3} = 8

— three presses, so

x = 3

, not

4

. Always translate the log into

b^{x} = y

before guessing.

lo g_{3} \frac{1}{9} = 2

since

3^{2} = 9

and

\frac{1}{9}

is about

9

The minus sign was dropped. We need

3^{x} = \frac{1}{9}

, and

3^{- 2} = \frac{1}{3 ^{2}} = \frac{1}{9}

, so the exponent is

- 2

. Getting below

1

with base

3 > 1

forces a negative exponent.

4^{x} = 32

x = lo g_{4} 32

, and since

32 > 4

x

must be a whole number."

Being bigger than the base does not force an integer. Rewrite in base

2

2^{2 x} = 2^{5}

, so

2 x = 5

and

x = \frac{5}{2}

. Non-integer answers are perfectly normal for logs.

lo g_{b} y

is defined whenever

y \neq = 0

, so

lo g_{2} (- 4)

is fine."

Wrong boundary. The requirement is

y > 0

, not merely

y \neq = 0

. Since

b^{x} > 0

always, its inverse can only accept positive inputs, ruling out every negative

y

"To undo

y = b^{x}

we just divide both sides by

b

Division cannot free a trapped exponent. Dividing by

b

gives

y / b = b^{x - 1}

— the exponent is still upstairs. Only the log, built specifically for this job, brings the exponent down.

lo g_{10} (2 \times 3) = lo g_{10} 2 \times lo g_{10} 3

Products of logs is not a rule. The product rule converts a product inside the log into a sum of logs:

lo g_{10} (2 \times 3) = lo g_{10} 2 + lo g_{10} 3

. See the log laws.

"The domain of

b^{x}

x > 0

, so the domain of

lo g_{b} x

is also

x > 0

The first half is already wrong:

b^{x}

accepts all real

x

. Its range is

y > 0

, and inverting swaps domain and range, so

lo g_{b}

has domain

x > 0

. The correct answer is right but the stated reason is broken.

Why questions

State the reason from the definition, not a memorised slogan.

Why must the base satisfy

b \neq = 1

Because

1^{x} = 1

for every

x

, the function

1^{x}

is a flat line — many inputs give the same output, so it is not one-to-one and has no inverse to name

lo g_{1}

Why must the base satisfy

b > 0

A negative base like

(- 2)^{x}

jumps between positive and negative values (and goes complex for fractional

x

), so it is not a smooth one-to-one function and cannot be cleanly inverted. See Exponential functions — $b^x$ and its graph.

Why is the range of

lo g_{b}

all real numbers?

The range of

lo g_{b}

equals the domain of

b^{x}

, and

b^{x}

accepts every real exponent. Inverting swaps domain and range, so the log outputs can be any real number.

Why does

y = lo g_{b} x

have a vertical asymptote while

y = b^{x}

has a horizontal one?

The log graph is the exponential graph reflected in

y = x

(see Inverse functions and reflection in $y=x$). Reflection swaps the axes, so the horizontal asymptote

y = 0

b^{x}

becomes the vertical asymptote

x = 0

lo g_{b} x

Why does every

lo g_{b} x

graph pass through

(1, 0)

Because

b^{0} = 1

for all valid

b

, we have

lo g_{b} 1 = 0

. Reflecting the point

(0, 1)

b^{x}

in the line

y = x

gives

(1, 0)

Why can we invent the symbol

lo g_{b}

instead of solving for

x

with algebra?

No combination of

+, -, \times, \div

can pull an exponent down from "upstairs". Since the operation genuinely doesn't exist among ordinary arithmetic, we name the inverse and call it

lo g_{b}

Why does converting

4^{x} = 32

to a common base of

2

work?

Once both sides read

2^{something}

, the equality

2^{2 x} = 2^{5}

forces the exponents equal (

2 x = 5

) because

2^{x}

is one-to-one. See Solving exponential equations.

If you can answer "why is $lo g_{b} 0$ undefined but $lo g_{b} 1 = 0$ ?" in a single breath, you own this topic. Because $b^{x}$ approaches $0$ without ever reaching it (no exponent lands on $0$ ), yet $b^{0}$ exactly equals $1$ (so the exponent giving $1$ is precisely $0$ ). :::

Connections

Exponential functions — $b^x$ and its graph
Laws of logarithms — product, quotient, power rules
Change of base formula
Natural logarithm $\ln$ and Euler's number $e$
Solving exponential equations
Inverse functions and reflection in $y=x$

Question bank — Logarithm — definition as inverse of exponential

True or false — justify

Spot the error

Why questions

Edge cases

Connections