3.2.12Exponentials & Logarithms

Graphs of logarithmic functions

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WHAT is a logarithmic function?

WHY a1a\neq 1 and a>0a>0? Because 1y=11^y=1 for every yy (no unique inverse), and negative bases give undefined values for many exponents. Same restriction as exponentials — logs inherit it.


HOW the graph is built (derivation from scratch)

We already know the exponential graph y=axy=a^x (for a>1a>1):

  • passes through (0,1)(0,1)
  • has a horizontal asymptote y=0y=0
  • always positive, increasing.

To get y=logaxy=\log_a x we reflect in y=xy=x (because that is what "inverse function" means geometrically — swap input and output). Reflecting a point (p,q)(p,q) in y=xy=x gives (q,p)(q,p). So every feature swaps its roles:

Feature of y=axy=a^x Swap xyx\leftrightarrow y Feature of y=logaxy=\log_a x
through (0,1)(0,1) through ==(1,0)(1,0)==
horizontal asymptote y=0y=0 vertical asymptote ==x=0x=0==
defined for all xx defined only for x>0x>0
range all y>0y>0 range all real yy
through (1,a)(1,a) through (a,1)(a,1)
Figure — Graphs of logarithmic functions

Shape depending on the base

WHY does it rise so slowly? To increase yy by 11, you must multiply xx by aa. To get y=6y=6 with base 1010 you need x=106=1000000x=10^6=1000000. Big input → small output. This is the whole reason logs are used to "compress" huge ranges (decibels, pH, Richter scale).


Transformations of log\log graphs

Using log laws we can rewrite transformed graphs as vertical shifts:

loga(kx)=logak+logax\log_a(kx) = \log_a k + \log_a x

So multiplying inside by a constant kk is the same as shifting the whole curve up by logak\log_a k. That's a neat, testable fact.

Standard transformations of y=logaxy=\log_a x:

  • y=logax+cy=\log_a x + c → shift up cc.
  • y=loga(xb)y=\log_a(x-b) → shift right bb, asymptote moves to x=bx=b.
  • y=logaxy=-\log_a x → reflect in xx-axis.

Worked examples


Common mistakes (steel-manned)


Recall Feynman: explain to a 12-year-old

A logarithm is a "how many times do I multiply" machine. log101000\log_{10}1000 asks "how many 10s multiplied make 1000?" → three. Now draw a picture of it: the graph starts way down low near the left edge (you can never touch the up-and-down line at x=0x=0), crosses the bottom at x=1x=1 (because log\log of 1 is always 0 — you multiply zero times to get 1), then climbs up but gets lazier and lazier, needing 10× more input for each extra step. It's literally the exponential graph looked at in a mirror placed along the diagonal line.


Active recall

What transformation maps y=axy=a^x onto y=logaxy=\log_a x?
Reflection in the line y=xy=x (inverse function).
Through which fixed point does every y=logaxy=\log_a x pass?
(1,0)(1,0), since loga1=0\log_a 1=0.
What is the asymptote of y=logaxy=\log_a x and its type?
x=0x=0, a vertical asymptote.
State the domain and range of y=logaxy=\log_a x (a>1a>1).
Domain x>0x>0; range all real numbers.
Why is there no yy-intercept for y=logaxy=\log_a x?
Because x=0x=0 is excluded (asymptote); log of 0 is undefined.
For a>1a>1, is logax\log_a x increasing or decreasing, and how fast?
Increasing, but slowly — xx must be multiplied by aa to raise yy by 1.
What point (other than (1,0)(1,0)) always lies on y=logaxy=\log_a x?
(a,1)(a,1), since logaa=1\log_a a=1.
Describe the graph of y=loga(xb)y=\log_a(x-b).
y=logaxy=\log_a x shifted right by bb; asymptote at x=bx=b.
Rewrite loga(ax)\log_a(ax) as a shift.
logax+1\log_a x + 1 (shift up 1), using loga(ax)=logaa+logax\log_a(ax)=\log_a a+\log_a x.
Solve log10x=3\log_{10}x=3 from the graph reasoning.
x=103=1000x=10^3=1000 (rewrite as x=ayx=a^y).

Connections

Concept Map

inverse is

means

produces

swap x and y

swap x and y

inherited by

has

shape set by

a>1 increasing

log_a kx equals log_a k plus log_a x

is a

y equals a^x exponential

y equals log_a x

Reflect in line y=x

Inverse function

a>0 and a not 1

Defined only x>0

Vertical asymptote x=0

Anchor points 1,0 and a,1

Base determines direction

Rises slowly, compresses range

Vertical shift via log laws

Transformations shift or reflect

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, logarithm basically exponential ka ulta (inverse) hai. Agar y=axy=a^x ka graph tumhe pata hai, to y=logaxy=\log_a x ka graph banane ke liye bas usko y=xy=x line mein mirror kar do. Jaha exponential (0,1)(0,1) se guzarta hai, waha log (1,0)(1,0) se guzrega — bas xx aur yy swap ho jaate hain. Yehi ek trick se poora graph samajh aa jaata hai.

Do cheezein hamesha yaad rakho: log sirf x>0x>0 ke liye define hota hai (negative ya zero ka log nahi hota), aur x=0x=0 pe ek vertical asymptote hoti hai — curve niche -\infty ki taraf bhaagta hai par yy-axis ko chhoota nahi. Exponential mein horizontal asymptote thi, mirror karne pe wo vertical ban gayi. Isliye log graph ka koi yy-intercept nahi hota, sirf xx-intercept (1,0)(1,0) hota hai.

Log dheere-dheere badhta hai kyunki yy ko 11 badhane ke liye xx ko aa se multiply karna padta hai. Isliye log101000000=6\log_{10}1000000=6 — bada input, chhota output. Yahi property real life mein kaam aati hai (pH, decibel, Richter scale) jaha bahut badi range ko compress karna hota hai.

Transformations mein sirf ek jaal hai: andar ka change horizontal hota hai, bahar ka vertical. log(x3)\log(x-3) matlab right shift 3 (asymptote x=3x=3 pe), jabki logx+3\log x + 3 matlab up shift 3. Aur ek smart baat — loga(ax)=logax+1\log_a(ax)=\log_a x+1, yaani base ka multiply andar, wo actually curve ko upar shift kar deta hai. Exam mein yeh point aksar poochha jaata hai.

Go deeper — visual, from zero

Test yourself — Exponentials & Logarithms

Connections