Solving logarithmic equations: Question bank — Maths deep dive

Before the questions, look at the picture that governs everything: the graph of $y = lo g_{b} x$ .

True or false — justify

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

is a valid law.

False. Logs turn products into sums, so

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

; there is no rule at all for the log of a sum.

lo g_{3} (2 x + 1) = lo g_{3} (x + 7)

then

2 x + 1 = x + 7

True. Both sides are single logs of the same base and log is one-to-one (strictly increasing), so equal outputs force equal arguments — provided both arguments are positive.

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

then

x + (x - 4) = 1

False. The right side is a plain number, not a log, so you cannot equate arguments. Combine to

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

first, then exponentiate to get

x (x - 4) = 5

lo g_{b} (N) = c

and

b^{c} = N

are the same statement.

True. They are two spellings of one relationship: the log asks "

b

to what power gives

N

?" and the exponential form is the answer written out.

Every algebraic solution of a log equation is a genuine solution.

False. Combining or squaring can invent "extraneous roots" that make some log argument

\leq 0

; those must be rejected because the original equation is undefined there.

lo g_{b} 1 = 0

for every allowed base

b

True.

b^{0} = 1

for any

b > 0, b \neq = 1

, so the power that gives

1

is always

0

, independent of the base — this is where every curve in the figure crosses the

x

-axis.

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

False. The left is a log squared (multiply the value by itself); the right is

2 lo g_{2} x

by the power law. They agree only in special cases, not as an identity.

lo g_{b} M = lo g_{c} M

with

M > 1

then

b = c

True. For fixed

M > 1

the value of

lo g_{b} M

strictly changes as the base changes, so equal logs of the same

M

force equal bases.

lo g_{b} 0

is just a very large negative number.

False. It is undefined: no finite power of a positive base ever reaches

0

, so

lo g_{b} 0

has no value at all (it is a limit, not a number).

Spot the error

Look at the side-by-side comparison of the three most common law-misuses before you attempt this section.

lo g_{2} (3 x - 1) = 4 \Rightarrow 3 x - 1 = 2 \cdot 4 = 8

Error: exponentiating is

2^{4}

, not

2 \times 4

. Correct is

3 x - 1 = 2^{4} = 16

, because

lo g_{b} N = c ⟺ N = b^{c}

, an exponent, not a product.

lo g_{5} x + lo g_{5} (x - 4) = 1 \Rightarrow lo g_{5} (2 x - 4) = 1

Error: they added the arguments instead of multiplying. The product law gives

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

, i.e.

lo g_{5} (x^{2} - 4 x)

, not

lo g_{5} (2 x - 4)

lo g (A + B) = lo g A + lo g B

, so I split it."

Error: no law expands the log of a sum. Only products, quotients and powers have laws;

lo g (A + B)

stays as it is.

"From

x^{2} - 4 x - 5 = 0

I get

x = 5

x = - 1

, so both solve

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

Error: no domain check.

x = - 1

makes

lo g_{5} (- 1)

undefined, so it is extraneous and rejected; only

x = 5

survives (needs

x > 4

lo g_{3} (2 x + 1) = lo g_{3} (x + 7) \Rightarrow 2 x + 1 = x + 7 \Rightarrow x = 6

, done."

The number is right but the write-up skips the domain check. You must confirm both arguments are positive:

2 (6) + 1 = 13 > 0

and

6 + 7 = 13 > 0

✓, otherwise the equating step wasn't licensed.

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

, so

x

can be any real number."

Error: the identity is fine but the domain isn't. The left side needs

x > 0

, so the equivalence only holds for

x > 0

; writing

lo g_{2} x^{2}

hides a domain that the original demanded.

(lo g_{2} x)^{2} - 3 lo g_{2} x + 2 = 0

, let

u = x

, then

u^{2} - 3 u + 2 = 0

Error: the wrong substitution. The equation is quadratic in the log, so set

u = lo g_{2} x

; then back-substitute

lo g_{2} x = 1

and

lo g_{2} x = 2

to reach

x

Why questions

The product law is not a coincidence — it is addition of exponents seen through the log. The figure below makes it visible as adding segment lengths.

Why does

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

Because

b^{(\cdot)}

and

lo g_{b} (\cdot)

are inverse functions (Inverse functions); each undoes the other, so applying one right after the other returns the original input

N

Why must the base satisfy

b > 0

and

b \neq = 1

b \leq 0

then

b^{c}

misbehaves (undefined or oscillating) for many

c

, and if

b = 1

then

1^{c} = 1

always, so

lo g_{1}

can't distinguish different arguments — the inverse would not exist.

Why can we equate arguments only when the log is one-to-one?

One-to-one means each output comes from exactly one input, so equal outputs guarantee equal inputs; without that property two different arguments could share a log value and equating would lose or gain solutions.

Why do we combine logs before exponentiating rather than after?

You can only "undo" a single log with the base. If two logs are added, exponentiating term-by-term is illegal, so the product law must first collapse them into one log.

Why does squaring or combining introduce extraneous roots?

Those operations can be valid on a wider domain than the original equation allows, so they admit candidate values where an original log argument is

\leq 0

— algebra keeps them but the original equation rejects them.

Why does the product law

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

hold at all?

Write

M = b^{x}, N = b^{y}

; then

M N = b^{x + y}

, so the log of the product is

x + y

— exactly the two individual logs added. It is the index law

b^{x} b^{y} = b^{x + y}

wearing a log costume, and the figure shows those exponents as stacked segment lengths.

Why is

lo g_{b}

of a negative number undefined for real logs?

Because

b^{c} > 0

for every real

c

when

b > 0

; a positive base raised to any real power never lands on a negative, so no power answers "

b

to what gives a negative?"

Why does the substitution

u = lo g_{b} x

turn a "log-quadratic" into an ordinary quadratic?

Because the equation only involves

lo g_{b} x

and its square, renaming that single quantity as

u

leaves a standard

u^{2} + \dots

that Quadratic equations by substitution handles directly.

Edge cases

The next figure shows why squaring can smuggle in a bad root: a negative argument, forbidden on the log's domain, gets folded into the positive side when you square.

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

, what is the true domain before solving?

You need

x > 0

and

x - 4 > 0

simultaneously, so the combined domain is

x > 4

; any candidate not exceeding

4

is dead on arrival.

Can

lo g_{b} (x - 3) = lo g_{b} (3 - x)

ever have a solution?

No. The two arguments can't both be positive at once (

x > 3

x < 3

), so no

x

lies in the domain of both sides — the equation is empty.

What happens to

lo g_{b} (N)

N \to 0^{+}

It plunges to

- \infty

; the log grows without bound in the negative direction as the argument approaches zero from above, which is why

0

itself is excluded (see the diving red curve in the first figure).

If a log equation yields

x = 4

but the domain requires

x > 4

, is

x = 4

a solution?

No. The boundary is open:

x = 4

makes an argument exactly

0

, which is undefined, so strict inequality means the endpoint is rejected.

Does

lo g_{b} x = lo g_{b} x

have a solution set, and what is it?

Yes — it holds for the whole domain

x > 0

(with

b > 0, b \neq = 1

), so it is an identity, not a single value; every positive

x

"solves" it.

Is there any

x

with

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

Only

x = 1

, since both sides equal

0

there; for

x > 1

(or

0 < x < 1

) the differing bases give differing values, so

x = 1

is the lone common point.

The whole workflow, as a flow-chart:

Recall One-line survival kit

Combine → Exponentiate → Check the domain. Most traps on this page are just the last step being skipped, or a law being applied to a sum instead of a product.

Connections

Laws of logarithms
Exponential equations
Change of base formula
Inverse functions
Quadratic equations by substitution
Domain and range of log functions