How do you solve #5 - 3^x = - 40#?

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Answer 1 · 2016-05-06T19:28:20

#x = 2+log_3(5)~~3.465#

#5-3^x = -40#

Subtract #5# from each side of the equation.

#-3^x = -45#

Multiply both sides of the equation by #-1#.

#3^x = 45#

Take the base-#3# logarithm of each side of the equation.

#log_3(3^x) = log_3(45)#

Apply the rule #log_a(a^x) = x# to the left hand side.

#x = log_3(45)#

Apply the rule #log(ab) = log(a)+log(b)# to the right hand side.

#x = log_3(9*5) = log_3(9)+log_3(5)#

Apply the rule #log_a(a^x) = x# to #log_3(9)=log_3(3^2)#

#x = 2+log_3(5)~~3.465#