How do you solve #3^(2x+1) = 5#?

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Answer 1 · 2017-08-03T16:33:09

#x = 0.232 # (to 3 sf)

We will not attempt to treat this as an exponential equation by working with the bases, because #5# is not a power of #3#.

Find the log of both sides.

#log 3^(2x+1) = log5" "larr# use the log power law

#(2x+1) log3 = log5" "larr# isolate the #x# term

#2x+1= log5/log3#

#2x = log5/log3-1#

#x = (log5/log3-1)div2#

There are no laws to simplify this, so use a calculator to find a value for #x#

#x = 0.23248676#