How do you solve #12^{4x}=4600#?

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Answer 1 · 2016-10-28T22:17:41

#x = 0.8485#

The first thought would be to write #4600# as a power of #12# so that the indices can be equated.

However, #12# has #3# as a factor, which #4600# does not. ( the sum of its digits is 10, which is not divisible by 3.)

Second thought: The variable is in the index - use logs

#12^(4x) = 4600#

#log 12^(4x) = log4600" "larr#log both sides

#4xlog12 = log4600" "larr# use the power law

#4x = (log4600)/(log12)" "larr# use a calculator

#4x = 3.3940#

#x = 0.8485" "larr#isolate #x#