Logarithms

What a logarithm is

A logarithm answers the question "what exponent do I need?" The statement $\log_b(x) = k$ means exactly the same thing as $b^k = x$. So $\log_5(125) = 3$ because $5^3 = 125$, and $\log_3(81) = 4$ because $3^4 = 81$. When no base is written, $\log$ means base 10.

The three properties

Three properties let logarithms be combined or split apart.

• Product rule: $\log(a) + \log(b) = \log(ab)$. A sum of logs becomes the log of a product.
• Quotient rule: $\log(a) - \log(b) = \log\!\left(\dfrac{a}{b}\right)$. A difference becomes the log of a quotient.
• Power rule: $\log(x^n) = n \cdot \log(x)$. An exponent comes out front as a multiplying coefficient.

The most common mistake is to treat these like ordinary arithmetic. $\log(6) + \log(4)$ is $\log(24)$, not $\log(10)$: adding logs multiplies the arguments, and subtracting logs divides them. Likewise $3\log(x)$ is $\log(x^3)$, not $\log(3x)$.

These rules apply only where the logarithms involved are defined. In the real-number setting used here, each logarithm's argument must be positive.

Using the properties with given values

Some questions hand you a couple of log values and ask for another. Break the new number into factors or powers of the numbers you were given. If $\log(2) \approx 0.301$ and $\log(3) \approx 0.477$, then $6 = 2 \times 3$, so $\log(6) = \log(2) + \log(3) \approx 0.778$. Similarly $8 = 2^3$, so $\log(8) = 3\log(2) \approx 0.903$. If $\log(4) = c$, then $64 = 4^3$ gives $\log(64) = 3c$. Always add for products and multiply by the exponent for powers. Multiplying two log values together is never the right move.

Solving a logarithmic equation

Solving a logarithmic equation means rewriting it in exponential form. $\log_7(x) = k$ is equivalent to $x = 7^k$: the base stays the base, the logarithm's value becomes the exponent, and the logarithm's argument becomes the exponential result. More generally, for any base $b$, $\log_b(x) = k$ is equivalent to $x = b^k$.

When an equation has two log terms, combine them into a single logarithm first using the product or quotient rule, then either set the arguments equal (if each side is a single log) or convert to exponential form.

Worked Example. Solve $\log(x) + \log(5) = \log(35)$.

Combine the left side using the product rule, then set the arguments equal:

$\log(5x) = \log(35)$

$5x = 35$

$x = 7$

Solving an exponential equation with logarithms

An exponential equation whose base does not convert neatly, such as $2^x = 30$, is solved by taking the logarithm of both sides: $\log(2^x) = \log(30)$, so $x \cdot \log(2) = \log(30)$, giving $x = \dfrac{\log(30)}{\log(2)}$. This works with any base of logarithm, as long as the same base is used in both the numerator and denominator.

Worked Example. Solve $4^x = 50$ using logarithms. Round to the nearest hundredth.

Take the log of both sides, then solve for $x$:

$x \cdot \log(4) = \log(50)$

$x = \dfrac{\log(50)}{\log(4)} \approx 2.82$

If the exponent is more than just $x$, finish the algebra after taking logs. For $3^{2x} = 50$, you get $2x = \dfrac{\log 50}{\log 3} \approx 3.56$, so $x \approx 1.78$. Stopping at $2x$ is an easy way to lose the point.

A full worked example: rejecting a bad solution

Worked Example. Solve $\log_2(x) + \log_2(x - 2) = 3$.

Recognize. Two logarithms with the same base are being added. Combine them with the product rule before converting to exponential form.

Set up. $\log_2\big(x(x-2)\big) = 3$.

Solve. $x(x-2) = 2^3 = 8$, so $x^2 - 2x - 8 = 0$, which factors as $(x-4)(x+2) = 0$. Then $x = 4$ or $x = -2$.

Check. $x = -2$ would require $\log_2(-2)$, which is undefined, so it is rejected. For $x = 4$: $\log_2(4) + \log_2(2) = 2 + 1 = 3$. ✓

Best route. Hand algebra after combining the logs. A calculator does not decide which candidate solutions are valid.

Common error. Reporting both $x = 4$ and $x = -2$ without checking that a logarithm's argument must be positive.

Where this appears as a model

Logarithms measure quantities on a log scale. A sound with 100 times the reference intensity has a decibel level of $10\log(100) = 20$ decibels, using the defining relationship between decibels and intensity ratio. In problems like this, a log turns a huge multiplicative range into a small, manageable number.

Strategy checklist

• Translate: $\log_b(x) = k$ means $b^k = x$. Say it out loud as "base to the answer gives the argument."
• Combine before solving: two logs of the same base become one log (product for $+$, quotient for $-$).
• Powers of the unknown: take logs of both sides and divide, then finish any remaining algebra.
• Check domains: plug each candidate back into the original equation and discard any that make an argument zero or negative.
• Common errors: adding arguments instead of multiplying, multiplying log values instead of adding, swapping base and exponent, and forgetting the last step of the algebra.