How do you solve #4 ^(2x) = 6#?

1 Answer
Noah G
Mar 28, 2016

You must use logarithms to solve.

Explanation:

If #a^n = b#, then #loga^n = logb#.

#log4^(2x) = log6#

Use the log rule #loga^n = nloga# to simplify further.

#2xlog4 = log6#

#x(log4^2) = log6#

#x = (log6)/(log16)#

#x = 0.65#

Make sure to ask your teacher if they want your answer in exact form (#x = (log6)/(log16)#) or rounded off to an x number of decimal points. I know that my teacher asks for two decimals after the point (#x = 0.65#), but some ask for 1, 3 or 0.

Practice exercises:

  1. Solve for x:

#a) 2^(3x - 2) = 17#

#b) 3^(4x + 1) = 6^(3x - 3)#

Challenge problem:

Solve for x in #2^(2x - 3) + 5 = 3^(3x)#

Hint: think of the log addition rule.

Good luck!

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