Proofs of the Base-10 Logarithm Laws

These workings derive the laws of base-10 logarithms from exponent laws by converting between exponential form and logarithmic form.

Assume a > 0, b > 0, and n ≠ 0.

Product Rule

log(ab) = log a + log b

Let 10x = a
Let 10y = b

Therefore:

log10a = x
log10b = y

10x10y = ab

10x+y = ab

Therefore:

log10(ab) = x + y

Substituting:

log10(ab) = log10a + log10b

Therefore:

log(ab) = log a + log b

Quotient Rule

log(a / b) = log a - log b

Let 10x = a
Let 10y = b

Therefore:

log10a = x
log10b = y

10x / 10y = a / b

10x-y = a / b

Therefore:

log10(a / b) = x - y

Substituting:

log10(a / b) = log10a - log10b

Therefore:

log(a / b) = log a - log b

Power Rule

log(an) = n log a

Let log(an) = x

Therefore:

10x = an

Taking the n-th root of both sides:

(10x)1/n = (an)1/n

10x/n = a

Therefore:

log10a = x / n

n log10a = x

Therefore:

log(an) = n log a

Root Rule

log(n√a) = (1 / n)log a

Let log(n√a) = x

Therefore:

10x = n√a

Since:

n√a = a1/n

Then:

10x = a1/n

Raising both sides to the power n:

(10x)n = (a1/n)n

10nx = a

Therefore:

log10a = nx

x = (1 / n)log10a

Therefore:

log(n√a) = (1 / n)log a

Common Logarithms and Natural Logarithms

log a = ln a × log e

Let ln a = y

Therefore:

ey = a

Let log10e = z

Therefore:

10z = e

Because ey = a:

(10z)y = a

10yz = a

Therefore:

log10a = yz

Substituting:

log10a = ln a × log10e

Therefore:

log a = ln a × log e

Summary

log(ab) = log a + log b

log(a / b) = log a - log b

log(an) = n log a

log(n√a) = (1 / n)log a

log a = ln a × log e

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