Week 6: Logarithmic Functions, Properties & Equations

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2026-09-12T00:00:00.000Z
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Logarithm definitions, core log laws, change of base, exponential & logarithmic equations, and domain traps. # Mathematics I — Week 6: Logarithmic Functions, Properties & Equations IITM’s Week 6 covers **logarithmic functions, their properties and graphs, exponential equations, and logarithmic equations**.

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Mathematics I — Week 6: Logarithmic Functions, Properties & Equations

IITM’s Week 6 covers logarithmic functions, their properties and graphs, exponential equations, and logarithmic equations.

TL;DR

A logarithm answers:
“What exponent gives this number?”
The fundamental equivalence is:
ax=y  ⟺  log⁡ay=x\boxed{ a^x=y \iff \log_a y=x }
with:
a>0,a≠1,y>0.a>0,\qquad a\neq1,\qquad y>0.
The highest-value Week 6 ideas are:
  • logs are inverses of exponentials
  • logarithm domain is positive reals
  • product / quotient / power laws
  • change of base
  • solving exponential equations
  • solving logarithmic equations
  • checking domain restrictions
  • graph relationship between axa^x and log⁡ax\log_a x

1. What is a logarithm?

Suppose:
23=8.2^3=8.
Then we can write:
log⁡28=3.\log_2 8=3.
Read:
logarithm base 22 of 88 equals 33
because:
23=8.2^3=8.
So logarithm asks:
To what power must the base be raised to obtain the given number?

2. Fundamental definition

For:
a>0,a≠1,a>0,\qquad a\neq1,
we define:
log⁡ax=y  ⟺  ay=x\boxed{ \log_a x=y \iff a^y=x }
This is the core of the entire week.

3. Examples

log⁡28=3\log_2 8=3
because:
23=8.2^3=8. log⁡10100=2\log_{10}100=2
because:
102=100.10^2=100. log⁡319=−2\log_3\frac19=-2
because:
3−2=19.3^{-2}=\frac19.

4. Why logarithms exist

Exponentials answer:
x↦ax.x\mapsto a^x.
Logarithms reverse that:
ax↦x.a^x\mapsto x.
So:
log⁡ax=(ax)−1\boxed{ \log_a x = (a^x)^{-1} }
in the inverse-function sense, not reciprocal sense.
Week 5's inverse-functions material leads directly here.

5. Basic restrictions

For real logarithms:
a>0,a≠1,x>0\boxed{ a>0,\quad a\neq1,\quad x>0 }
Why?
The base must be positive and not equal to 11.
And the logarithm's argument must be positive.
So:
log⁡a0\log_a 0
is undefined over the reals.
And:
log⁡a(−5)\log_a(-5)
is also undefined over the reals.

6. Domain and range

For:
f(x)=log⁡ax,f(x)=\log_a x,
the domain is:
(0,∞)\boxed{ (0,\infty) }
and the range is:
R\boxed{ \mathbb R }
Compare with exponential:
ax:Domain=R,Range=(0,∞).a^x: \quad \text{Domain}=\mathbb R, \quad \text{Range}=(0,\infty).
The domain/range swap happens because logarithm is the inverse.

7. Basic log values

Since:
a0=1,a^0=1,
we get:
log⁡a1=0\boxed{ \log_a1=0 }
Since:
a1=a,a^1=a,
we get:
log⁡aa=1\boxed{ \log_a a=1 }
These should become automatic.

8. Inverse identities

Because exponential and logarithm undo each other:
log⁡a(ax)=x\boxed{ \log_a(a^x)=x }
and:
alog⁡ax=x\boxed{ a^{\log_a x}=x }
provided the expressions are valid.

9. Graph of logarithm

For:
a>1,a>1,
the graph of:
y=log⁡axy=\log_a x
is increasing.
It passes through:
(1,0)(1,0)
because:
log⁡a1=0.\log_a1=0.
It also passes through:
(a,1).(a,1).

10. Exponential/log reflection

The graphs of:
y=axy=a^x
and:
y=log⁡axy=\log_a x
are reflections across:
y=x.\boxed{y=x}.
Why?
They are inverse functions.
So points swap:
(x,y)↔(y,x).(x,y) \leftrightarrow (y,x).

11. Vertical asymptote

For:
y=log⁡ax,y=\log_a x,
the yy-axis:
x=0x=0
is a vertical asymptote.
As:
x→0+,x\to0^+,
for a>1a>1:
log⁡ax→−∞.\log_a x\to-\infty.
The function never reaches or crosses:
x=0.x=0.

12. Logarithm growth vs decay

If:
a>1,a>1,
then:
log⁡ax\log_a x
is increasing.
If:
0<a<1,0<a<1,
then:
log⁡ax\log_a x
is decreasing.
This mirrors exponential growth/decay.

13. Product law

One of the most important properties:
log⁡a(xy)=log⁡ax+log⁡ay\boxed{ \log_a(xy) = \log_a x+\log_a y }
Why?
Suppose:
x=am,y=an.x=a^m,\qquad y=a^n.
Then:
xy=am+n.xy=a^{m+n}.
So:
log⁡a(xy)=m+n.\log_a(xy)=m+n.
But:
m=log⁡ax,n=log⁡ay.m=\log_a x,\qquad n=\log_a y.
Therefore:
log⁡a(xy)=log⁡ax+log⁡ay.\log_a(xy)=\log_a x+\log_a y.

14. Quotient law

log⁡a(xy)=log⁡ax−log⁡ay\boxed{ \log_a\left(\frac xy\right) = \log_a x-\log_a y }
because exponent division subtracts powers:
aman=am−n.\frac{a^m}{a^n}=a^{m-n}.

15. Power law

log⁡a(xr)=rlog⁡ax\boxed{ \log_a(x^r) = r\log_a x }
because:
x=amx=a^m
implies:
xr=amr.x^r=a^{mr}.
So logarithms turn exponents into multiplication.
This is one reason they are so useful.

16. Important false identities

These are wrong:
log⁡(x+y)≠log⁡x+log⁡y\log(x+y) \neq \log x+\log y
and:
log⁡(x−y)≠log⁡x−log⁡y.\log(x-y) \neq \log x-\log y.
The laws apply to:
  • multiplication
  • division
  • powers
not ordinary addition/subtraction inside the argument.

17. Compressing logarithms

Suppose:
log⁡ax+log⁡ay.\log_a x+\log_a y.
Use product law:
log⁡a(xy)\boxed{ \log_a(xy) }
Suppose:
2log⁡ax−log⁡ay.2\log_a x-\log_a y.
Power law:
2log⁡ax=log⁡a(x2).2\log_a x=\log_a(x^2).
Then quotient law:
log⁡a(x2y)\boxed{ \log_a\left(\frac{x^2}{y}\right) }

18. Expanding logarithms

Example:
log⁡a(x3y2z).\log_a\left(\frac{x^3y^2}{z}\right).
Apply quotient:
=log⁡a(x3y2)−log⁡az.= \log_a(x^3y^2)-\log_a z.
Product:
=log⁡ax3+log⁡ay2−log⁡az.= \log_a x^3+\log_a y^2-\log_a z.
Power law:
3log⁡ax+2log⁡ay−log⁡az\boxed{ 3\log_a x+2\log_a y-\log_a z }

19. Change-of-base formula

For any valid base bb:
log⁡ax=log⁡bxlog⁡ba\boxed{ \log_a x = \frac{\log_b x}{\log_b a} }
This lets us compute logarithms in arbitrary bases using a calculator that only knows, say:
ln⁡\ln
or:
log⁡10.\log_{10}.

20. Deriving change of base

Let:
y=log⁡ax.y=\log_a x.
Then:
ay=x.a^y=x.
Take log⁡b\log_b of both sides:
log⁡b(ay)=log⁡bx.\log_b(a^y)=\log_b x.
Power law:
ylog⁡ba=log⁡bx.y\log_b a=\log_b x.
So:
y=log⁡bxlog⁡ba.y= \frac{\log_b x}{\log_b a}.
Therefore:
log⁡ax=log⁡bxlog⁡ba\boxed{ \log_a x = \frac{\log_b x}{\log_b a} }

21. Common logarithm

When the base is:
10,10,
we often write:
log⁡x\log x
instead of:
log⁡10x.\log_{10}x.
This is called the common logarithm.

22. Natural logarithm

When the base is:
e,e,
we write:
ln⁡x\boxed{ \ln x }
instead of:
log⁡ex.\log_e x.
Here:
e≈2.71828.e\approx2.71828.
This base becomes extremely important in calculus, probability, statistics, growth processes and machine learning.

23. Why ee is special

You don't need the full calculus story yet.
For now, know:
exe^x
is the natural exponential function.
Its inverse is:
ln⁡x.\ln x.
So:
ln⁡(ex)=x\ln(e^x)=x
and:
eln⁡x=x.e^{\ln x}=x.

24. Solving exponential equations: same base

Example:
2x+1=16.2^{x+1}=16.
Write:
16=24.16=2^4.
Then:
2x+1=24.2^{x+1}=2^4.
Since exponential functions are one-one:
x+1=4.x+1=4.
Thus:
x=3\boxed{x=3}

25. Another same-base example

9x=27.9^x=27.
Rewrite:
9=329=3^2
and:
27=33.27=3^3.
Therefore:
(32)x=33.(3^2)^x=3^3.
So:
32x=33.3^{2x}=3^3.
Hence:
2x=32x=3
and:
x=32\boxed{x=\frac32}

26. Exponential equations without same base

Suppose:
2x=7.2^x=7.
We cannot rewrite 77 as a neat integer power of 22.
Take logarithms:
log⁡(2x)=log⁡7.\log(2^x)=\log7.
Power law:
xlog⁡2=log⁡7.x\log2=\log7.
Thus:
x=log⁡7log⁡2\boxed{ x= \frac{\log7}{\log2} }
which is:
log⁡27.\log_2 7.

27. General exponential equation

For:
ax=b,a^x=b,
we get:
x=log⁡ab\boxed{ x=\log_a b }
or via natural logs:
x=ln⁡bln⁡a\boxed{ x= \frac{\ln b}{\ln a} }

28. Exponential equation with coefficient

Solve:
3⋅2x=24.3\cdot2^x=24.
First isolate exponential:
2x=8.2^x=8.
Then:
x=3.x=3.
Always isolate the exponential part before taking logs when possible.

29. Example with exponent expression

Solve:
52x−1=11.5^{2x-1}=11.
Take logs:
ln⁡(52x−1)=ln⁡11.\ln(5^{2x-1})=\ln11.
Power law:
(2x−1)ln⁡5=ln⁡11.(2x-1)\ln5=\ln11.
Therefore:
2x−1=ln⁡11ln⁡5.2x-1=\frac{\ln11}{\ln5}.
So:
x=12(1+ln⁡11ln⁡5)\boxed{ x= \frac12 \left( 1+\frac{\ln11}{\ln5} \right) }

30. Solving logarithmic equations

Example:
log⁡2x=5.\log_2 x=5.
Convert to exponential form:
x=25.x=2^5.
Thus:
x=32\boxed{x=32}
This is often the fastest method.

31. Example

Solve:
log⁡3(x−1)=2.\log_3(x-1)=2.
Convert:
x−1=32.x-1=3^2.
So:
x−1=9x-1=9
and:
x=10\boxed{x=10}
Check domain:
x−1>0.x-1>0.
1010 satisfies it.

32. Log equation using product law

Solve:
log⁡2x+log⁡2(x−2)=3.\log_2 x+\log_2(x-2)=3.
First, domain:
x>0x>0
and:
x−2>0.x-2>0.
Therefore:
x>2.x>2.
Combine logs:
log⁡2[x(x−2)]=3.\log_2[x(x-2)]=3.
Convert to exponential form:
x(x−2)=23=8.x(x-2)=2^3=8.
Thus:
x2−2x−8=0.x^2-2x-8=0.
Factor:
(x−4)(x+2)=0.(x-4)(x+2)=0.
Candidates:
x=4,−2.x=4,\quad -2.
But domain requires:
x>2.x>2.
Therefore:
x=4\boxed{x=4}

33. Extraneous solutions matter

Logarithmic equations frequently produce algebraic candidates that violate the original domain.
So always check:
every logarithm argument must be >0\boxed{ \text{every logarithm argument must be }>0 }
This is not optional.

34. Example with rejected root

Solve:
log⁡(x−1)+log⁡(x−3)=log⁡3.\log(x-1)+\log(x-3)=\log3.
Domain:
x>3.x>3.
Combine:
log⁡[(x−1)(x−3)]=log⁡3.\log[(x-1)(x-3)]=\log3.
Assuming same valid base:
(x−1)(x−3)=3.(x-1)(x-3)=3.
Expand:
x2−4x+3=3.x^2-4x+3=3.
So:
x2−4x=0.x^2-4x=0.
Thus:
x(x−4)=0.x(x-4)=0.
Candidates:
x=0,4.x=0,\quad4.
But:
x>3.x>3.
Therefore:
x=4\boxed{x=4}

35. Equality of logarithms

If:
log⁡au=log⁡av\log_a u=\log_a v
with valid arguments and valid base, then:
u=v\boxed{u=v}
because logarithmic functions are one-one.
This is the log analogue of:
au=av⇒u=v.a^u=a^v\Rightarrow u=v.

36. Example

Solve:
log⁡5(x+2)=log⁡5(3x−4).\log_5(x+2)=\log_5(3x-4).
Set arguments equal:
x+2=3x−4.x+2=3x-4.
Thus:
6=2x6=2x
and:
x=3.x=3.
Check:
x+2=5>0x+2=5>0
and:
3x−4=5>0.3x-4=5>0.
So valid.

37. Graph transformations

Start with:
y=log⁡ax.y=\log_a x.
Then:
y=log⁡a(x−h)+ky=\log_a(x-h)+k
shifts:
  • right by hh
  • up by kk
The vertical asymptote moves from:
x=0x=0
to:
x=h.\boxed{x=h}.

38. Example

f(x)=log⁡2(x−3)+5.f(x)=\log_2(x-3)+5.
Domain requires:
x−3>0.x-3>0.
So:
x>3\boxed{x>3}
Vertical asymptote:
x=3\boxed{x=3}
Range:
R.\mathbb R.

39. Finding domain of logarithmic expressions

Example:
f(x)=log⁡(x+4).f(x)=\log(x+4).
Require:
x+4>0.x+4>0.
Thus:
x>−4\boxed{x>-4}

40. Multiple factors inside log domain

Suppose:
f(x)=log⁡[(x−1)(x+3)].f(x)=\log[(x-1)(x+3)].
Require:
(x−1)(x+3)>0.(x-1)(x+3)>0.
Critical points:
x=−3,1.x=-3,\quad1.
Product is positive outside:
x<−3orx>1\boxed{ x<-3 \quad\text{or}\quad x>1 }
Notice:
Each factor does not individually need to be positive.
The entire logarithm argument must be positive.
That distinction matters.

41. Logarithm of reciprocal

Using quotient law:
log⁡a(1x)=log⁡a1−log⁡ax.\log_a\left(\frac1x\right) = \log_a1-\log_a x.
Since:
log⁡a1=0,\log_a1=0,
we get:
log⁡a(1x)=−log⁡ax\boxed{ \log_a\left(\frac1x\right) = -\log_a x }

42. Negative exponents and logs

Because:
a−x=1ax,a^{-x}=\frac1{a^x},
we get:
log⁡a(a−x)=−x.\log_a(a^{-x})=-x.
This is another way exponent and log rules mirror each other.

43. Log of roots

Since:
xn=x1/n,\sqrt[n]{x}=x^{1/n},
power law gives:
log⁡axn=1nlog⁡ax\boxed{ \log_a\sqrt[n]{x} = \frac1n\log_a x }
Example:
log⁡28=12log⁡28=32.\log_2\sqrt8 = \frac12\log_2 8 = \frac32.

44. Exponential growth model

A common model is:
P(t)=P0atP(t)=P_0a^t
or:
P(t)=P0ekt.P(t)=P_0e^{kt}.
Where:
  • P0P_0 = initial amount
  • tt = time
  • kk = growth/decay parameter
Logs allow us to solve for time.

45. Example: solving for time

Suppose:
P(t)=100(2)t.P(t)=100(2)^t.
When does:
P(t)=800?P(t)=800?
Solve:
100(2)t=800.100(2)^t=800.
Then:
2t=8.2^t=8.
So:
t=3\boxed{t=3}

46. General growth-time equation

Suppose:
P(t)=P0ekt.P(t)=P_0e^{kt}.
Then:
P(t)P0=ekt.\frac{P(t)}{P_0}=e^{kt}.
Take natural logs:
ln⁡(P(t)P0)=kt.\ln\left(\frac{P(t)}{P_0}\right)=kt.
Hence:
t=1kln⁡(P(t)P0)\boxed{ t= \frac1k \ln\left(\frac{P(t)}{P_0}\right) }
This kind of manipulation appears everywhere later.

47. Why logs are useful conceptually

Logarithms transform:

multiplication → addition

log⁡(xy)=log⁡x+log⁡y\log(xy)=\log x+\log y

division → subtraction

log⁡(x/y)=log⁡x−log⁡y\log(x/y)=\log x-\log y

exponentiation → multiplication

log⁡(xr)=rlog⁡x\log(x^r)=r\log x
So logs turn certain nonlinear operations into simpler algebraic ones.
Historically, this made hand computation dramatically easier.
Today it remains mathematically powerful.

48. Orders of magnitude

A base-1010 logarithm gives a rough measure of scale.
For example:
103=100010^3=1000
so:
log⁡10(1000)=3.\log_{10}(1000)=3.
And:
106=1,000,00010^6=1,000,000
so:
log⁡10(1,000,000)=6.\log_{10}(1,000,000)=6.
A difference of 33 in base-1010 logs represents a factor of:
103=1000.10^3=1000.
This is why logarithmic scales are useful for quantities spanning huge ranges.

Common mistake: log of zero
log⁡a0\log_a0
is not:
0.0.
It is undefined.
Remember:
log⁡a1=0.\log_a1=0.
Common mistake: confusing
log⁡a1\log_a1 and log⁡1a\log_1a
log⁡a1=0\log_a1=0
for valid base aa.
But:
log⁡1a\log_1a
is invalid because logarithm base:
11
is not allowed.

Common mistake: product law backwards
Correct:
log⁡(xy)=log⁡x+log⁡y.\log(xy)=\log x+\log y.
Wrong:
log⁡(x+y)=log⁡x+log⁡y.\log(x+y)=\log x+\log y.
There is generally no simple decomposition for:
log⁡(x+y).\log(x+y).
Common mistake: power law
Correct:
log⁡(x3)=3log⁡x.\log(x^3)=3\log x.
But be careful with the domain over real numbers.
Expressions such as:
log⁡(x2)\log(x^2)
may exist for negative xx even though:
log⁡x\log x
does not.
So transformations involving logs must respect domains.
For standard course problems, check the original expression.

Common mistake: forgetting base conditions
The base must satisfy:
a>0,a≠1\boxed{ a>0,\qquad a\neq1 }
A base of 00, negative number, or 11 does not define the ordinary real logarithm function used here.

Common mistake: solving before domain check
Given:
log⁡(x−5)=log⁡(2−x),\log(x-5)=\log(2-x),
before manipulating anything:
x−5>0⇒x>5x-5>0 \Rightarrow x>5
and:
2−x>0⇒x<2.2-x>0 \Rightarrow x<2.
No xx satisfies both.
Therefore the equation has:
no real solution\boxed{\text{no real solution}}
You can sometimes solve a question immediately from the domain.

55. A useful solving workflow

For a logarithmic equation:
text
1. Write domain restrictions
2. Simplify/merge logarithms
3. Convert log form ↔ exponential form
4. Solve resulting algebra
5. Check every candidate against original domain
For exponential equations:
text
1. Isolate exponential term
2. Try rewriting both sides with same base
3. If impossible, take logs
4. Use power law
5. Solve algebraically

56. Full example: exponential equation

Solve:
4x+1=7.4^{x+1}=7.
Take natural logs:
ln⁡(4x+1)=ln⁡7.\ln(4^{x+1})=\ln7.
Power law:
(x+1)ln⁡4=ln⁡7.(x+1)\ln4=\ln7.
Thus:
x+1=ln⁡7ln⁡4.x+1=\frac{\ln7}{\ln4}.
Therefore:
x=ln⁡7ln⁡4−1\boxed{ x= \frac{\ln7}{\ln4}-1 }

57. Full example: logarithmic equation

Solve:
log⁡3(x+1)+log⁡3(x−1)=2.\log_3(x+1)+\log_3(x-1)=2.

Domain

x+1>0x+1>0
and:
x−1>0.x-1>0.
So:
x>1.x>1.

Combine

log⁡3[(x+1)(x−1)]=2.\log_3[(x+1)(x-1)]=2.

Convert

(x+1)(x−1)=32.(x+1)(x-1)=3^2.
So:
x2−1=9.x^2-1=9.
Thus:
x2=10.x^2=10.
Candidates:
x=±10.x=\pm\sqrt{10}.
Domain requires:
x>1.x>1.
Therefore:
x=10\boxed{x=\sqrt{10}}

58. Exponential/log symmetry table

ExponentialLogarithm
y=axy=a^xy=log⁡axy=\log_a x
Domain R\mathbb RDomain (0,∞)(0,\infty)
Range (0,∞)(0,\infty)Range R\mathbb R
passes (0,1)(0,1)passes (1,0)(1,0)
horizontal asymptote y=0y=0vertical asymptote x=0x=0
a0=1a^0=1log⁡a1=0\log_a1=0
a1=aa^1=alog⁡aa=1\log_a a=1
They are mirror images conceptually and geometrically.

59. Formula sheet

Definition

log⁡ax=y  ⟺  ay=x\boxed{ \log_a x=y \iff a^y=x }

Product

log⁡a(xy)=log⁡ax+log⁡ay\boxed{ \log_a(xy)=\log_a x+\log_a y }

Quotient

log⁡axy=log⁡ax−log⁡ay\boxed{ \log_a\frac xy = \log_a x-\log_a y }

Power

log⁡a(xr)=rlog⁡ax\boxed{ \log_a(x^r)=r\log_a x }

Change of base

log⁡ax=log⁡bxlog⁡ba\boxed{ \log_a x = \frac{\log_bx}{\log_ba} }

Inverse identities

log⁡a(ax)=x\boxed{ \log_a(a^x)=x } alog⁡ax=x\boxed{ a^{\log_a x}=x }

60. Weeks 5–6 as one idea

Week 5:
f(x)=axf(x)=a^x
Week 6:
f−1(x)=log⁡ax.f^{-1}(x)=\log_a x.
So don't mentally store these as unrelated chapters.
They are one reversible relationship:
x⟶axax⟶log⁡ax\boxed{ x \overset{a^x}{\longrightarrow} a^x \overset{\log_a}{\longrightarrow} x }

61. What you should actually master

By the end of Week 6, you should be able to:
  1. Translate between: ay=xa^y=x and: log⁡ax=y.\log_a x=y.
  2. Know log domain/base restrictions.
  3. Use:
    • product law
    • quotient law
    • power law.
  4. Expand and condense logarithmic expressions.
  5. Apply change of base.
  6. Recognize exponential/log graphs as inverses.
  7. Identify domain, range and asymptotes.
  8. Solve same-base exponential equations.
  9. Solve arbitrary-base exponential equations using logs.
  10. Solve basic logarithmic equations.
  11. Combine multiple logarithms before solving.
  12. Reject extraneous/domain-invalid answers.
  13. Work with ln⁡\ln and ee at a basic level.
If those are comfortable, Week 6 is under control.

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Week 7 moves into sequences and limits, functions of one variable, graphs/tangents, limits of sequences, and limits of functions.
That is the start of the calculus portion of Maths 1, and the conceptual shift is important:
finite algebra→behavior as something approaches a value\boxed{ \text{finite algebra} \rightarrow \text{behavior as something approaches a value} }

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Quick Recall & Flashcards

Test your active recall before proceeding to practice problems. Format:
Question :: Answer.
  • Definition of log⁡b(x)=y\log_b(x) = y :: by=xb^y = x (x>0,b>0,b≠1x > 0, b > 0, b \neq 1)
  • Product rule for logarithms: log⁡b(xy)\log_b(xy) :: log⁡b(x)+log⁡b(y)\log_b(x) + \log_b(y)
  • Quotient rule for logarithms: log⁡b(x/y)\log_b(x/y) :: log⁡b(x)−log⁡b(y)\log_b(x) - \log_b(y)
  • Power rule for logarithms: log⁡b(xk)\log_b(x^k) :: klog⁡b(x)k \log_b(x)
  • Change of base formula for log⁡b(a)\log_b(a) :: ln⁡(a)ln⁡(b)=log⁡10(a)log⁡10(b)\frac{\ln(a)}{\ln(b)} = \frac{\log_{10}(a)}{\log_{10}(b)}
  • Natural logarithm ln⁡(x)\ln(x) base :: Base e≈2.71828e \approx 2.71828
  • Dangerous trap when solving log equations :: Must check for extraneous roots (log argument must be >0> 0)

Practice & Self-Check

Attempt these diagnostic questions to verify your operational understanding.
1

Question 1

Solve for xx: log⁡2(x)+log⁡2(x−2)=3\log_2(x) + \log_2(x - 2) = 3.
NAT
3 Unit Assessment
2

Question 2

What is the value of log⁡3(81)\log_3(81)?
NAT
3 Unit Assessment

Active Recall Session

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