# How do you solve log2x=log4?

Mar 1, 2016

Put the logs to one side and solve using the rule ${\log}_{a} m - {\log}_{a} n = {\log}_{a} \left(\frac{m}{n}\right)$

#### Explanation:

$0 = \log 4 - \log 2 x$

$0 = \log \left(\frac{4}{2 x}\right)$

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

${10}^{0} = \frac{4}{2 x}$

$1 = \frac{4}{2 x}$

$2 x = 4$

$x = 2$

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 ${\log}_{a} m - {\log}_{a} n = {\log}_{a} \left(\frac{m}{n}\right)$ and ${\log}_{a} m + {\log}_{a} n = {\log}_{a} \left(m \times n\right)$ are extremely important when working with logarithms.

Practice exercises:

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

a) $\log 3 x = \log 12$

b) $\log \left(x + 1\right) + \log \left(4\right) = 0$

c) ${\log}_{5} \left(2 x + 3\right) = {\log}_{5} \left(3 x + 1\right) + {\log}_{5} \left(7\right)$

Good luck!