Question

How do you solve `log2x=log4`?

Answer 1

Put the logs to one side and solve using the rule `log_am - log_an = log_a(m/n)`

Explanation:

`0 = log4 - log2x`

`0 = log(4/(2x))`

Convert to exponential form. The log is in base 10 because nothing is written in subscript to the right of the log.

`10^0 = 4/(2x)`

`1 = 4/(2x)`

`2x = 4`

`x = 2`

You could have done this much easier by just dividing 4 by 2, since the logs are in the same base. However, I did the long way because that method won't work when you have logarithmic equations like b) and c) in the practice exercises.

Note that the rules `log_am - log_an = log_a(m/n)` and `log_am + log_an = log_a(m xx n)` are extremely important when working with logarithms.

Practice exercises:

1. Solve for x. Round answers to two decimals.

a) `log3x = log12`

b) `log(x + 1) + log(4) = 0`

c) `log_5(2x + 3) = log_5(3x + 1) + log_5(7)`

Good luck!