How do you solve #2( 7) ^ { 4x - 5} = 30#?

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Answer 1 · 2017-07-22T00:53:49

#x ~~ 1.598#

We're asked to solve the equation

#2(7)^(4x-5) = 30#

for #x#.

The first thing we can do is divide both sides by #2#:

#7^(4x-5) = 15#

Change this to logarithmic form:

#4x - 15 = log_(7)15#

Change-of-base formula gives

#4x-5 = (log15)/(log7)#

#4x = (log15)/(log7) + 5#

#x = color(red)((log15)/(4log7) + 5/4# #~~ color(red)(1.598#