Logarithms (Algebra II) · lesson

A log is an exponent

💡 log_b x asks: b to what power gives x?

log₄ 64 = 3 because 4³ = 64. The answer to a log is always an exponent.

log with no base written means base 10. ln means base e.

Fractions and decimals give negative logs: log₂ (1/8) = −3, because 2^(−3) = 1/8.

🧠 Let's work it out

Find log 0.01.

What do you think comes next?

💡 Adding logs with the same base multiplies the insides, and subtracting divides them.

log_b M + log_b N = log_b (MN). log_b M − log_b N = log_b (M/N).

Combine first, then ask what power you need. Neither log alone may be a whole number, but the combined one often is.

🧠 Let's work it out

Find log₃ 54 − log₃ 2.

What do you think comes next?

💡 To solve b^x = c, take the log of both sides: x = log c ÷ log b, or x = ln c when the base is e.

First get the power alone. If the equation is 300 · 1.05^t = 600, divide by 300 to get 1.05^t = 2.

Then x = log c ÷ log b. Either log key works, as long as you use the same one on top and bottom.

For e^x = c, ln undoes e, so x = ln c.

🧠 Let's work it out

Solve 3^x = 50.

What do you think comes next?

💡 Rewrite log_b A = c as b^c = A, and throw out any answer that takes the log of a negative number or zero.

log₅ (x − 1) = 2 means 5² = x − 1, so x = 26. The same idea finds a missing base: log_b 64 = 3 means b³ = 64, so b = 4.

If there are two logs, combine them into one with the log rules first.

You can only take the log of a positive number. Check each answer in the original equation.

🧠 Let's work it out

Solve log₃ x + log₃ (x − 8) = 2.

What do you think comes next?

extraneous solution: an answer from the algebra that doesn't work in the original equation