Log of 1 and Log of the Base

Evaluate $\log_9 1 + \log_9 9$.

Evaluate $\log_9 1$: Which power of $9$ gives $1$? $9^0 = 1$ = $\log_9 1 = 0$

Evaluate $\log_9 9$: Which power of $9$ gives $9$? $9^1 = 9$ = $\log_9 9 = 1$

Add the results: $0 + 1$ = $1$

Answer: $\log_9 1 + \log_9 9 = 1$

Log of a Power of the Base

Evaluate $\log_4 64$ without a calculator.

Write the argument as a power of the base: $4 \times 4 \times 4 = 64$ = $64 = 4^3$

Rewrite the logarithm: $\log_4 64 = \log_4(4^3)$ = $\log_4(4^3)$

Use $\log_b(b^x) = x$: The logarithm and the power cancel, the exponent remains = $3$

Answer: $\log_4 64 = 3$

Base Raised to a Logarithm

Simplify $3^{\log_3 7} + 2^{\log_2 5}$.

Check the bases: In $3^{\log_3 7}$ both bases are $3$, in $2^{\log_2 5}$ both bases are $2$ = The bases match

Use $b^{\log_b x} = x$ on the first term: $\log_3 7$ is the exponent that turns $3$ into $7$ = $3^{\log_3 7} = 7$

Use the same property on the second term: $\log_2 5$ is the exponent that turns $2$ into $5$ = $2^{\log_2 5} = 5$

Add: $7 + 5$ = $12$

Answer: $3^{\log_3 7} + 2^{\log_2 5} = 12$

Using the One-to-One Property

Solve $\log_3(2x + 1) = \log_3(x + 6)$.

Same base on both sides, so set the arguments equal: If $\log_3 M = \log_3 N$, then $M = N$ = $2x + 1 = x + 6$

Solve the linear equation: $2x - x = 6 - 1$ = $x = 5$

Check that both arguments are positive: $2 \cdot 5 + 1 = 11 > 0$ and $5 + 6 = 11 > 0$ = Both arguments are $11$

Answer: $x = 5$

One-to-One Property With a Domain Check

Solve $\log_4(x^2 - 4) = \log_4(3x)$.

Set the arguments equal: One-to-one property with base $4$ = $x^2 - 4 = 3x$

Write the quadratic in standard form: $x^2 - 3x - 4 = 0$ = $(x - 4)(x + 1) = 0$

Find both candidates: $x - 4 = 0$ or $x + 1 = 0$ = $x = 4$ or $x = -1$

Check $x = -1$: $3x = -3$, and $\log_4(-3)$ is undefined = Reject $x = -1$

Check $x = 4$: $4^2 - 4 = 12 > 0$ and $3 \cdot 4 = 12 > 0$ = $x = 4$ works

Answer: $x = 4$ (the candidate $x = -1$ is rejected)

Mistake: Writing $\log_b 1 = 1$ or $\log_b 1 = b$

Why: The logarithm is the exponent, not the result. The question is: which power of $b$ gives $1$?

Correct: $b^0 = 1$, so $\log_b 1 = 0$ for every allowed base. For example $\log_7 1 = 0$.

Mistake: Thinking $\log_b 0 = 0$ or that the log of a negative number is a negative number

Why: A positive base raised to any power is always positive, so no exponent gives $0$ or a negative number.

Correct: $\log_b 0$ and $\log_b(-4)$ are undefined. The argument must be greater than $0$, while the value of a logarithm can be negative: $\log_2 \frac{1}{8} = -3$.

Mistake: Cancelling a power and a logarithm with different bases: $2^{\log_3 9} = 9$

Why: $b^{\log_b x} = x$ only works when the base of the power and the base of the logarithm are the same number.

Correct: Here the bases differ: $\log_3 9 = 2$, so $2^{\log_3 9} = 2^2 = 4$, not $9$.

Mistake: Using the one-to-one property and not checking the arguments

Why: Setting the arguments equal can produce a value that makes an argument zero or negative, where the original logarithm does not exist.

Correct: Always put each candidate back into every argument. Keep it only if all arguments are positive.

Sound Level in Decibels

The loudness of a sound is $L = 10 \log \frac{I}{I_0}$ decibels, where $I_0$ is the quietest sound a person can hear. The basic properties make the calculation quick.

At the hearing threshold $I = I_0$, so $L = 10 \log 1 = 10 \cdot 0 = 0$ decibels.

Acidity and pH

Chemists measure acidity with $\text{pH} = -\log[H^+]$, where $[H^+]$ is the hydrogen ion concentration in moles per liter. The inverse property turns a pH back into a concentration.

Pure water has $[H^+] = 10^{-7}$, so $\text{pH} = -\log(10^{-7}) = -(-7) = 7$.

Rounds in a Knockout Tournament

In a knockout tournament half of the teams drop out in every round. With $2^n$ teams there are $\log_2(2^n) = n$ rounds until one winner is left.

With $16 = 2^4$ teams there are $\log_2 16 = \log_2(2^4) = 4$ rounds.

Q: Why is $\log_b 1$ always $0$?

A: Any allowed base raised to the power $0$ equals $1$: $2^0 = 1$, $10^0 = 1$, $e^0 = 1$. The logarithm asks for that exponent, so the answer is always $0$.

Q: Can the value of a logarithm be negative?

A: Yes. Only the argument has to be positive. For example $\log_{10} 0.01 = -2$ because $10^{-2} = 0.01$. The argument $0.01$ is positive, the value $-2$ is negative.

Q: Why does $b^{\log_b x} = x$ need $x > 0$, while $\log_b(b^x) = x$ works for every $x$?

A: In $\log_b(b^x)$ the argument $b^x$ is always positive, so every real $x$ is allowed. In $b^{\log_b x}$ the logarithm $\log_b x$ must exist first, and that needs $x > 0$.

Q: Where are the product, quotient and power rules?

A: Each has its own lesson: Product Rule for Logarithms, Quotient Rule for Logarithms and Power Rule for Logarithms. They use the properties from this lesson, so it helps to master these first.