How do you solve #log2x=log4#?

1 Answer
Noah G
Mar 1, 2016

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!

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