Solving exponential equations using logarithms: Question bank

This page assumes only the parent method: to free an exponent you take a log of both sides and use the power law $lo g_{b} (m^{k}) = k lo g_{b} m$ . Everything else is built here.

True or false — justify

You can take any log base (10,

e

, base 7) and still get the same final answer for

x

True —

x = \frac{l o g c}{l o g b}

is a ratio of two logs in the same base, and that ratio is base-independent. See Change of Base Formula: every base gives the identical number.

lo g (m + n) = lo g m + lo g n

False. Logs turn products into sums:

lo g (mn) = lo g m + lo g n

. There is no clean rule for the log of a sum — you cannot split

lo g (m + n)

at all.

To solve

5 \cdot 3^{x} = 60

you should start by writing

lo g (5 \cdot 3^{x}) = lo g 60

Technically legal but foolish. First divide by 5 to isolate the power (

3^{x} = 12

); otherwise you carry an extra

lo g 5

term for no reason. Isolate, then log.

\frac{lo g c}{lo g b}

equals

lo g (\frac{c}{b})

False — a classic trap. The left side is a ratio of two logs (

= lo g_{b} c

); the right is a difference

lo g c - lo g b

. Different operations, different values.

b^{x} = c

and

c < 0

, there is still a real solution for

x

False. For a positive base

b > 1

b^{x}

is always positive for every real

x

, so it can never equal a negative

c

. No real solution — see Inverse Functions (log of a negative is undefined).

The power law

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

requires

k

to be a whole number.

False. The derivation (

b^{y} = m \Rightarrow b^{y k} = m^{k}

) never assumed

k

was an integer, so

k

can be any real — fractions, decimals, negatives all work.

2^{x + 1} = 8^{x}

is best solved by taking logs of both sides.

False — it's possible but wasteful. Since

8 = 2^{3}

, rewrite as

2^{x + 1} = 2^{3 x}

and match exponents:

x + 1 = 3 x

. Matching a common base gives an exact answer with no decimals.

Applying

lo g

to both sides of a true equation keeps it true.

True, provided both sides are positive (log's domain). Since

b^{x} > 0

and we only solve for

c > 0

, the sides are valid inputs and equality is preserved — it's a legal "do the same thing to both sides" move.

Spot the error

7^{2 x - 1} = 40 \Rightarrow 2 x lo g 7 - 1 = lo g 40

The bracket was dropped. The whole exponent comes down:

(2 x - 1) lo g 7 = lo g 40

. The

- 1

is inside the exponent, so it must be multiplied by

lo g 7

, not left stranded.

lo g (3^{x}) = x \cdot lo g 3

, therefore

lo g (3 + x) = lo g 3 + lo g x

The first is the power law (correct). The second is invented — there is no rule for

lo g

of a sum. The student confused an exponent coming down with a fictional "distribute over addition".

3^{x} = 12

, so

x = lo g 12 - lo g 3 = lo g 4

Wrong tool. That subtraction gives

lo g (12/3) = lo g 4

, which is a plain number, not

x

. Correct is the ratio

x = \frac{l o g 12}{l o g 3} = lo g_{3} 12

. Division of logs, not subtraction.

4^{x} = 1

has no solution because you can't take

lo g

of both sides usefully."

You can — and

x = 0

. Note

\frac{l o g 1}{l o g 4} = \frac{0}{l o g 4} = 0

. Any positive base to the power

0

1

, so

x = 0

is the clean answer.

"To solve

2 \cdot 5^{x} = 5^{x + 1}

, take logs immediately."

Better to simplify first:

5^{x + 1} = 5 \cdot 5^{x}

, so the equation is

2 \cdot 5^{x} = 5 \cdot 5^{x}

, i.e.

2 = 5

after dividing by

5^{x}

. That's false, so there is no solution — logs would only obscure this.

x = \frac{l o g 7}{l o g 2}

, so

lo g 2

divides into

lo g 7

leaving

lo g 5

Nonsense algebra. You cannot "cancel" or subtract inside a ratio of logs.

\frac{l o g 7}{l o g 2}

is just

lo g_{2} 7 \approx 2.807

— leave it as a single division.

Why questions

Why must we bring the exponent down rather than take a root to solve

b^{x} = c

A root undoes a fixed power (

x^{2}

), but here the unknown is the power itself. The log is the operation matched to that structure — it converts an unknown exponent into an unknown multiplier we can divide off.

Why do we divide off any coefficient (like the 5 in

5 \cdot 3^{x}

) before logging?

The power law only pulls down an exponent when the exponential term stands alone. A coefficient multiplies the whole term, not the exponent, so logging early just adds a stray

lo g 5

you'd have to subtract later anyway.

Why does the final answer

x = \frac{l o g c}{l o g b}

not depend on which log button you press?

Changing base multiplies top and bottom by the same factor, which cancels. Formally this ratio equals

lo g_{b} c

regardless of intermediate base — the Change of Base Formula guarantees it.

Why is matching bases (Example-4 style) considered "smarter" than logging?

When both sides are powers of one base, exponents can be equated directly, giving an exact rational answer (like

x = \frac{1}{2}

) instead of a rounded decimal from a log ratio.

Why can log undo exponentials at all?

Because they are inverse functions:

lo g_{b}

is defined precisely as the function that reverses

b^{()}

, just as

reverses squaring. Applying inverse-then-original returns the input. See Inverse Functions.

Why is

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

the only log law the whole method really needs?

Every exponential equation's job is to free the exponent; this single law converts "

x

upstairs" into "

x

as a multiplier". The product/quotient rules (Laws of Logarithms) only help tidy coefficients, not free the exponent.

Edge cases

What is the solution of

b^{x} = 1

for any valid base

b > 0, b \neq = 1

Always

x = 0

, since

b^{0} = 1

by definition. Via the formula,

\frac{l o g 1}{l o g b} = \frac{0}{l o g b} = 0

— the numerator is zero.

What happens if you try

1^{x} = 5

No solution. Base

1

gives

1^{x} = 1

for every

x

, so it can never equal

5

. This is why the base is required to satisfy

b \neq = 1

before the method applies.

Does

b^{x} = c

with

0 < b < 1

(a decaying base) still work with the same formula?

Yes.

x = \frac{l o g c}{l o g b}

holds, but

lo g b < 0

, so dividing flips signs — a decaying base gives a negative

x

for

c > 1

, matching Exponential Growth and Decay.

c = b

, what does the formula predict, and does it make sense?

x = \frac{l o g b}{l o g b} = 1

, and indeed

b^{1} = b

. The self-consistent check: the ratio of a log with itself is

1

Can

x

be negative in

b^{x} = c

, and when?

Yes, whenever

0 < c < 1

with base

b > 1

: e.g.

2^{x} = \frac{1}{2}

gives

x = - 1

. A fraction on the right forces a negative exponent, since

b^{- n} = 1/ b^{n}

What if both sides can't be matched to a common base, like

2^{x} = 3^{x + 1}

You must log (matching fails). Log gives

x lo g 2 = (x + 1) lo g 3

; expand and collect the

x

terms — it reduces to ordinary linear algebra, just with log constants.

Graphically, why does

b^{x} = c

have exactly one solution when

b > 1, c > 0

The curve

y = b^{x}

is strictly increasing and covers all positive

y

, so the horizontal line

y = c

crosses it once and only once — see Solving Equations Graphically.

Recall One-line self-test

If you can state why $\frac{l o g c}{l o g b}$ is a ratio not a difference, why you isolate before logging, and when there is no real solution — you own this topic.

Connections

Laws of Logarithms — product/quotient/power rules the traps exploit.
Change of Base Formula — why the answer is base-independent.
Inverse Functions — why log undoes exp, and why $lo g$ of negatives fails.
The Number e and Natural Logarithms — same traps hold with $ln$ .
Exponential Growth and Decay — where negative/decaying-base cases arise.
Solving Equations Graphically — the one-intersection uniqueness argument.